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Full text of "Siyavula textbooks: Grade 12 Physical Science"

Siyavula textbooks: Grade 12 Physical 

Science 



Collection Editor: 

Free High School Science Texts Project 



Siyavula textbooks: Grade 12 Physical 

Science 



Collection Editor: 

Free High School Science Texts Project 

Authors: 

Free High School Science Texts Project 

Rory Adams 

Heather Williams 

Kosma von Maltitz 



Online: 

< http://siyavula.cnx.0rg/content/colli244/l.2/ > 



CONNEXIONS 
Rice University, Houston, Texas 



This selection and arrangement of content as a collection is copyrighted by Free High School Science Texts Project. 

It is licensed under the Creative Commons Attribution 3.0 license (http://creativecommons.Org/licenses/by/3.0/). 

Collection structure revised: August 3, 2011 

PDF generated: August 3, 2011 

For copyright and attribution information for the modules contained in this collection, see p. 325. 



Table of Contents 



1 Organic molecules 

1.1 Introduction and key concepts 1 

1.2 Hydrocarbons 7 

1.3 The alcohols 19 

1.4 Carboxylic acids, amines and carboxlyic acid derivatives 21 

Solutions 29 

2 Organic macromolecules 

2.1 Polymers 33 

2.2 Biological macromolecules 43 

Solutions 52 

3 Reaction rates 

3.1 Introduction 53 

3.2 Collision theory, measurement and mechanism 58 

3.3 Chemical equilibrium 64 

3.4 The equilibrium constant 67 

3.5 Le Chateliers principle 70 

Solutions 78 

4 Electrochemical reactions 

4.1 The galvanic cell 81 

4.2 The electrolytic cell 87 

4.3 Standard potentials 89 

4.4 Balancing redox reactions 96 

4.5 Applications 97 

Solutions 103 

5 The chemical industry 

5.1 Sasol 107 

5.2 The chloralkaH industry 113 

5.3 The fertilizer industry 118 

5.4 Electrochemistry and batteries 123 

6 Motion in two dimensions 

6.1 Vertical projectile motion 133 

6.2 Conservation of momentum 137 

6.3 Types of collisions 139 

6.4 Frames of reference 144 

Solutions 155 

7 Mechanical properties of matter 

7.1 Deformation of materials 171 

7.2 Failure and strength of materials 175 

Solutions 179 

8 Work, energy and power 

8.1 Work 181 

8.2 Energy 185 

8.3 Power 190 

Solutions 195 

9 Doppler Effect 201 



IV 

10 Colour 

10.1 Colour and light 209 

10.2 Paints and pigments 214 

Solutions 217 

11 2D and 3D wavefronts 

11.1 Wavefronts, Huygen's principle and interference 219 

11.2 Diffraction 222 

11.3 Shock waves and sonic booms 225 

Solutions 232 

12 Wave nature of matter 

12.1 de Broglie wavelength 235 

12.2 Electron microscopes 237 

Solutions 240 

13 Electrodynamics 

13.1 Generators and motors 243 

13.2 Alternating current, inductance and capacitance 248 

Solutions 253 

14 Electronics 

14.1 Capacitive and inductive circuits 255 

14.2 Principles of digital electronics 260 

14.3 Active circuit elements 271 

Solutions 281 

15 Electromagnetic radiation 

15.1 Wave nature and particle nature 283 

15.2 Penetrating ability 287 

Solutions 290 

16 Optical phenomena and properties of matter 

16.1 Transmission and scattering of light 291 

16.2 Photoelectric effect 294 

16.3 Emission and absorption spectra 298 

16.4 Lasers 303 

Solutions 312 

Glossary 315 

Index 322 

Attributions 325 



Chapter 1 

Organic molecules 

1.1 Introduction and key concepts' 

1.1.1 What is organic chemistry? 

Organic chemistry is the branch of chemistry that deals with organic molecules. An organic molecule 
is one which contains carbon, and these molecules can range in size from simple molecules to complex 
structures containing thousands of atoms! Although carbon is present in all organic compounds, other 
elements such as hydrogen (H), oxygen (O), nitrogen (N), sulfur (S) and phosphorus (P) are also common 
in these molecules. 

Until the early nineteenth century, chemists had managed to make many simple compounds in the 
laboratory, but were still unable to produce the complex molecules that they found in living organisms. 
It was around this time that a Swedish chemist called Jons Jakob Berzelius suggested that compounds 
found only in living organisms (the organic compounds) should be grouped separately from those found in 
the non-living world (the inorganic compounds). He also suggested that the laws that governed how organic 
compounds formed, were different from those for inorganic compounds. From this, the idea developed that 
there was a 'vital force' in organic compounds. In other words, scientists believed that organic compounds 
would not follow the normal physical and chemical laws that applied to other inorganic compounds because 
the very 'force of life' made them different. 

This idea of a mystical 'vital force' in organic compounds was weakened when scientists began to man- 
ufacture organic compounds in the laboratory from non-living materials. One of the first to do this was 
Friedrich Wohler in 1828, who successfully prepared urea, an organic compound in the urine of animals 
which, until that point, had only been found in animals. A few years later a student of Wohler's, Hermann 
Kolbe, made the organic compound acetic acid from inorganic compounds. By this stage it was acknowl- 
edged that organic compounds are governed by exactly the same laws that apply to inorganic compounds. 
The properties of organic compounds are not due to a 'vital force' but to the unique properties of the carbon 
atom itself. 

Organic compounds are very important in daily life. They make up a big part of our own bodies, they 
are in the food we eat and in the clothes we wear. Organic compounds are also used to make products such 
as medicines, plastics, washing powders, dyes, along with a list of other items. 

1.1.2 Sources of carbon 

The main source of the carbon in organic compounds is carbon dioxide in the air. Plants use sunlight to 
convert carbon dioxide into organic compounds through the process of photosynthesis. Plants are therefore 



^This content is available online at <http://siyavula.cnx.Org/content/m39444/l.l/>. 



2 CHAPTER 1. ORGANIC MOLECULES 

able to make their own organic compounds through photosynthesis, while animals feed on plants or plant 
products so that they gain the organic compounds that they need to survive. 

Another important source of carbon is fossil fuels such as coal, petroleum and natural gas. This is 
because fossil fuels are themselves formed from the decaying remains of dead organisms (refer to Grade 11 
for more information on fossil fuels). 

1,1,3 Unique properties of carbon 

Carbon has a number of unique properties which influence how it behaves and how it bonds with other 
atoms: 

• Carbon has four valence electrons which means that each carbon atom can form four bonds with 
other atoms. Because of this, long chain structures can form. These chains can either be unbranched 
(Figure 1.1) or branched (Figure 1.2). Because of the number of bonds that carbon can form with 
other atoms, organic compounds can be very complex. 

Image notjtnished 

Figure 1.1: An unbranched carbon chain 



Image notjinished 

Figure 1.2: A branched carbon chain 



Because of its position on the Periodic Table, most of the bonds that carbon forms with other atoms are 
covalent. Think for example of a C-C bond. The difference in electronegativity between the two atoms 
is zero, so this is a pure covalent bond. In the case of a C-H bond, the difference in electronegativity 
between carbon (2.5) and hydrogen (2.1) is so small that C-H bonds are almost purely covalent. The 
result of this is that most organic compounds are non-polar. This affects some of the properties of 
organic compounds. 



1,1,4 Representing organic compounds 

There are a number of ways to represent organic compounds. It is useful to know all of these so that you 
can recognise a molecule however it is shown. There are three main ways of representing a compound. We 
will use the example of a molecule called 2-methylpropane to help explain the difference between each. 

1.1.4.1 Molecular formula 

The molecular formula of a compound shows how many atoms of each type are in a molecule. The number 
of each atom is written as a subscript after the atomic symbol. The molecular formula of 2-methylpropane 

is: 

C4H10 



1.1.4.2 Structural formula 

The structural formula of an organic compound shows every bond between every atom in the molecule. Each 
bond is represented by a line. The structural formula of 2-methylpropane is shown in Figure 1.3. 

Image notjinished 

Figure 1.3: The structural formula of 2-methylpropane 

1.1.4.3 Condensed structural formula 

When a compound is represented using its condensed structural formula, each carbon atom and the hydrogen 
atoms that are bonded directly to it are listed as a molecular formula, followed by a similar molecular formula 
for the neighbouring carbon atom. Branched groups are shown in brackets after the carbon atom to which 
they are bonded. The condensed structural formula below shows that in 2-methylpropane, there is a branched 
chain attached to the second carbon atom of the main chain. You can check this by looking at the structural 
formula in . 

CH3CH(CH3)CH3 

1.1.4.3.1 Representing organic compounds 

1. For each of the following organic compounds, give the condensed structural formula and the 
molecular formula. 



a. 



Image notjinished 

Figure 1.4 

b Image notjinished 

Figure 1.5 



2. For each of the following, give the structural formula and the molecular formula. 

a. CH3CH2CH3 

b. CH3CH2CH(CH3)CH3 

C. C2H6 

3. Give two possible structural formulae for the compound with a molecular formula of C4H10. 



4 CHAPTER 1. ORGANIC MOLECULES 

1.1.5 Isomerism in organic compounds 

It is possible for two organic compounds to have the same molecular formula but a different structural 
formula. Look for example at the two organic compounds that are shown in Figure 1.6. 



H 




H 



H c H 



H 



H H H H 



H H 



H 



H 



H 



Figure 1.6: Isomers of a 4-carbon organic compound 



If you were to count the number of carbon and hydrogen atoms in each compound, you would find that 
they are the same. They both have the same molecular formula (C4H10), but their structure is different and 
so are their properties. Such compounds are called isomers. 

Definition 1.1: Isomer 

In chemistry, isomers are molecules with the same molecular formula and often with the same 
kinds of chemical bonds between atoms, but in which the atoms are arranged differently. 



1.1.5.1 Isomers 

Match the organic compound in Column A with its isomer Column B: 



Column A 


Column B 


CH3CH(CH3)OH 


CH3CH(CH 


Image notjinished 

Figure 1.7 




Image notjinished 

Figure 1.9 


C3H7OH 



Table 1.1 



1.1.6 Functional groups 

All organic compounds have a particular bond or group of atoms which we call its functional group. This 
group is important in determining how a compound will react. 

Definition 1.2: Functional group 

In organic chemistry, a functional group is a specific group of atoms within molecules, that are 
responsible for the characteristic chemical reactions of those molecules. The same functional group 
will undergo the same or similar chemical reaction(s) regardless of the size of the molecule it is a 
part of. 

In one group of organic compounds called the hydrocarbons, the single, double and triple bonds of the 
alkanes, alkenes and alkynes are examples of functional groups. In another group, the alcohols, an oxygen 
and a hydrogen atom are bonded to each other to form the functional group for those compounds (in other 
words an alcohol has an OH in it). All alcohols will contain an oxygen and a hydrogen atom bonded together 
in some part of the molecule. 

Table 1.2 summarises some of the common functional groups. We will look at these in more detail later 
in this chapter. 



Name of group 



Functional group 



Example 



Diagram 



continued on next page 



CHAPTER 1. ORGANIC MOLECULES 



Alkane 


Image not finished 

Figure 1.10 


Alkene 


Image notjinished 

Figure 1.12 


Alkyne 


Image notjinished 

Figure 1.14 


Halo-alkane 


Image notjinished 

Figure 1.16 


continued on next page 



Alcohoi/ alkanoi 


Image not finished 

Figure 1.18 


Carboxylic acid 


Image notjinished 

Figure 1.20 


Amine 


Image notjinished 

Figure 1.22 



Table 1.2: Some functional groups of organic compounds 



1.2 Hydrocarbons' 



1,2,1 The Hydrocarbons 

Let us first look at a group of organic compounds known as the hydrocarbons. These molecules only contain 
carbon and hydrogen. The hydrocarbons that we are going to look at are called aliphatic compounds. 
The aliphatic compounds are divided into acyclic compounds (chain structures) and cyclic compounds (ring 
structures). The chain structures are further divided into structures that contain only single bonds (alkanes), 
those that contain at least one double bond (alkenes) and those that contain at least one triple bond 
(alkynes). Cyclic compounds include structures such as the benzene ring. Figure 1.24 summarises the 
classification of the hydrocarbons. 



^This content is available online at <http://siyavula.cnx.Org/content/m39453/l.l/>. 



CHAPTER 1. ORGANIC MOLECULES 



Image notjinished 

Figure 1.24: The classification of the aliphatic hydrocarbons 



Hydrocarbons that contain only single bonds are called saturated hydrocarbons because each carbon 
atom is bonded to as many hydrogen atoms as possible. Figure 1.25 shows a molecule of ethane which is a 
saturated hydrocarbon. 



Image notjinished 

Figure 1.25: A saturated hydrocarbon 



Hydrocarbons that contain double or triple bonds are called unsaturated hydrocarbons because they 
don't contain as many hydrogen atoms as possible. Figure 1.26 shows a molecule of ethene which is an 
unsaturated hydrocarbon. If you compare the number of carbon and hydrogen atoms in a molecule of 
ethane and a molecule of ethene, you will see that the number of hydrogen atoms in ethene is less than the 
number of hydrogen atoms in ethane despite the fact that they both contain two carbon atoms. In order for 
an unsaturated compound to become saturated, a double bond has to be broken, and another two hydrogen 
atoms added for each double bond that is replaced by a single bond. 



Image notjinished 

Figure 1.26: An unsaturated hydrocarbon 



NOTE: Fat that occurs naturally in living matter such as animals and plants is used as food for 
human consumption and contains varying proportions of saturated and unsaturated fat. Foods that 
contain a high proportion of saturated fat are butter, ghee, suet, tallow, lard, coconut oil, cottonseed 
oil, and palm kernel oil, dairy products (especially cream and cheese), meat, and some prepared 
foods. Diets high in saturated fat are correlated with an increased incidence of atherosclerosis and 
coronary heart disease according to a number of studies. Vegetable oils contain unsaturated fats 
and can be hardened to form margarine by adding hydrogen on to some of the carbon=carbon 
double bonds using a nickel catalyst. The process is called hydrogenation 

We will now go on to look at each of the hydrocarbon groups in more detail. These groups are the alkanes, 
the alkenes and the alkynes. 



1.2.1.1 The Alkanes 

The alkanes are hydrocarbons that only contain single covalent bonds between their carbon atoms. This 
means that they are saturated compounds and are quite unreactive. The simplest alkane has only one carbon 
atom and is called methane. This molecule is shown in Figure 1.27. 

Image notjinished 

Figure 1.27: The structural (a) and molecular formula (b) for methane 
The second alkane in the series has two carbon atoms and is called ethane. This is shown in Figure 1.28. 

Image notjinished 

Figure 1.28: The structural (a) and molecular formula (b) for ethane 
The third alkane in the series has three carbon atoms and is called propane (Figure 1.29). 

Image notjinished 

Figure 1.29: The structural (a) and molecular formula (b) for propane 



When you look at the molecular formula for each of the alkanes, you should notice a pattern developing. 
For each carbon atom that is added to the molecule, two hydrogen atoms are added. In other words, each 
molecule differs from the one before it by CH2. This is called a homologous series. The alkanes have the 
general formula C„H2n+2- 

The alkanes are the most important source of fuel in the world and are used extensively in the chemical 
industry. Some are gases (e.g. methane and ethane), while others are liquid fuels (e.g. octane, an important 
component of petrol). 

NOTE: Some fungi use alkanes as a source of carbon and energy. One fungus Amorphotheca 
resinae prefers the alkanes used in aviation fuel, and this can cause problems for aircraft in tropical 
areas! 

1.2.1.2 Naming the alkanes 

In order to give compounds a name, certain rules must be followed. When naming organic compounds, the 
lUPAC (International Union of Pure and Applied Chemistry) nomenclature is used. We will first look at 



10 



CHAPTER 1. ORGANIC MOLECULES 



some of the steps that need to be followed when naming a compound, and then try to apply these rules to 
some specific examples. 

1. STEP 1: Recognise the functional group in the compound. This will determine the sufRx (the 'end') 
of the name. For example, if the compound is an alkane, the sufRx will be -ane; if the compound is an 
alkene the sufRx will be -ene; if the compound is an alcohol the sufRx will be -ol, and so on. 

2. STEP 2: Find the longest continuous carbon chain (it won't always be a straight chain) and count 
the number of carbon atoms in this chain. This number will determine the preRx (the 'beginning') of 
the compound's name. These preRxes are shown in Table 1.3. So, for example, an alkane that has 3 
carbon atoms will have the sufRx prop and the compound's name will be propane. 



Carbon atoms 


preRx 


1 


meth(ane) 


2 


eth(ane) 


3 


prop (ane) 


4 


but (ane) 


5 


pent (ane) 


6 


hex(ane) 


7 


hept(ane) 


8 


oct(ane) 


9 


non(ane) 


10 


dec (ane) 



Table 1.3: The preRx of a compound's name is determined by the number of carbon atoms in the 

longest chain 

3. STEP 3: Number the carbons in the longest carbon chain (Important: If there is a double or triple 
bond, you need to start numbering so that the bond is at the carbon with the lowest number. 

4. STEP 4: Look for any branched groups and name them. Also give them a number to show their 
position on the carbon chain. If there are no branched groups, this step can be left out. 

5. STEP 5: Combine the elements of the name into a single word in the following order: branched groups; 
preRx; name ending according to the functional group and its position along the longest carbon chain. 

Khan academy video on alkanes 

This media object is a Flash object. Please view or download it at 
<http://www.youtube.com/v/NRFPvLp3r3g&rel=0&hl=en_US&feature=player_embedded&version=3> 



Figure 1.30 



Exercise 1.1: Naming the alkanes 

Give the lUPAC name for the following compound: 



(Solution on p. 29.) 



11 



Image notjinished 



Figure 1.31 



Note: The numbers attached to the carbon atoms would not normally be shown. The atoms 
have been numbered to help you to name the compound. 

Exercise 1.2: Naming the alkanes (Solution on p. 29.) 

Give the lUPAC name for the following compound: 



Image notjinished 

Figure 1.32 



Exercise 1.3: Naming the alkanes (Solution on p. 29.) 

Give the lUPAC name for the following compound: 

CH3CH(CH3)CH(CH3)CH3 

(Remember that the side groups are shown in brackets after the carbon atom to which they are 
attached.) 

Exercise 1.4: Naming the alkanes (Solution on p. 30.) 

Give the lUPAC name for the following compound: 



Image notjinished 

Figure 1.33 



1.2.1.2.1 Naming the alkanes 

1. Give the structural formula for each of the following: 

a. Octane 

b. CH3CH2CH3 

c. CH3CH(CH3)CH3 

d. 3-ethyl-pentane 

2. Give the lUPAC name for each of the following organic compounds. 



Image notjinished 

Figure 1.34 



12 



CHAPTER 1. ORGANIC MOLECULES 



b. CH3CH2CH(CH3)CH2CH3 

c. CH3CH(CH3)CH2CH(CH3)CH3 

1.2.1.3 Properties of the alkanes 

We have aheady mentioned that the alkanes are relatively unreactive because of their stable C-C and C-H 
bonds. The boiling point and melting point of these molecules is determined by their molecular structure, 
and their surface area. The more carbon atoms there are in an alkane, the greater the surface area and 
therefore the higher the boiling point. The melting point also increases as the number of carbon atoms in 
the molecule increases. This can be seen in the data in Table 1.4. 



Formula 


Name 


Melting point 

("C) 


Boiling point 

(°C) 


Phase at room 
temperature 


CH4 


methane 


-183 


-162 


gas 


C2H6 


ethane 


-182 


-88 


gas 


CsHg 


propane 


-187 


-45 


gas 


C4H10 


butane 


-138 


-0.5 


gas 


C5H12 


pentane 


-130 


36 


liquid 


C6H14 


hexane 


-95 


69 


liquid 


C17H36 


heptadecane 


22 


302 


soHd 



Table 1.4: Properties of some of the alkanes 

You will also notice that, when the molecular mass of the alkanes is low (i.e. there are few carbon 
atoms), the organic compounds are gases because the intermolecular forces are weak. As the number of 
carbon atoms and the molecular mass increases, the compounds are more likely to be liquids or solids 
because the intermolecular forces are stronger. 

1.2.1.4 Reactions of the alkanes 

There are three types of reactions that can occur in saturated compounds such as the alkanes. 

1. Substitution reactions Substitution reactions involve the removal of a hydrogen atom which is 
replaced by an atom of another element, such as a halogen (F, CI, Br or I) (Figure 1.35). The product 
is called a halo-alkane. Since alkanes are not very reactive, either heat or light is needed for this 
reaction to take place, e.g. CH2=CH2 + HBr -^ CH3-CH2-Br (halo-alkane) 



H 



H 



H 



CI 



CL 



light 



■> 



H 



H 



H 



Figure 1.35: A substitution reaction 



13 

Halo-alkanes (also sometimes called alkyl halides) that contain methane and chlorine are substances 
that can be used as anaesthetics during operations. One example is trichloromethane, also known as 
'chloroform' (Figure 1.36). 

Image notjinished 

Figure 1.36: Trichloromethane 



2. Elimination reactions Saturated compounds can also undergo elimination reactions to become unsat- 
urated (Figure 1.37). In the example below, an atom of hydrogen and chlorine are eliminated from the 
original compound to form an unsaturated halo-alkene. e.g. CIH2C—CH2CI -^ H^C = CHCI + HCl 



Image notjinished 

Figure 1.37: An elimination reaction 



3. Oxidation reactions When alkanes are burnt in air, they react with the oxygen in air and heat is 
produced. This is called an oxidation or combustion reaction. Carbon dioxide and water are given off 
as products. Heat is also released during the reaction. The burning of alkanes provides most of the 
energy that is used by man. e.g. CH4 + 2O2 -^ CO2 + 2H2O + heat 



1.2.1.4.1 The Alkanes 



1. Give the lUPAC name for each of the following alkanes: 
a. CH3CH2CH2CH2CH2CH3 



Image notjinished 



b. 

Figure 1.38 



C. CH3CH3 

2. Give the structural formula for each of the following compounds: 

a. octane 

b. 3-methyl-hexane 

3. Methane is one of the simplest alkanes and yet it is an important fuel source. Methane occurs naturally 
in wetlands, natural gas and permafrost. However, methane can also be produced when organic wastes 
(e.g. animal manure and decaying material) are broken down by bacteria under conditions that are 
anaerobic (there is no oxygen). The simplified reaction is shown below: Organic matter -^ Simple 
organic acids -^ Biogas The organic matter could be carbohydrates, proteins or fats which are broken 
down by acid-forming bacteria into simple organic acids such as acetic acid or formic acid. Methane- 
forming bacteria then convert these acids into biogases such as methane and ammonia. The production 
of methane in this way is very important because methane can be used as a fuel source. One of the 



14 CHAPTER 1. ORGANIC MOLECULES 

advantages of methane over other fuels like coal, is that it produces more energy but with lower carbon 
dioxide emissions. The problem however, is that methane itself is a greenhouse gas and has a much 
higher global warming potential than carbon dioxide. So, producing methane may in fact have an even 
more dangerous impact on the environment. 

a. What is the structural formula of methane? 

b. Write an equation to show the reaction that takes place when methane is burned as a fuel. 

c. Explain what is meant by the statement that methane 'has a greater global warming potential 
than carbon dioxide'. 

4. Chlorine and ethane react to form chloroethane and hydrogen chloride. 

a. Write a balanced chemical equation for this reaction, using molecular formulae. 

b. Give the structural formula of chloroethane. 

c. What type of reaction has taken place in this example? 

5. Petrol (CgHis) is in fact not pure CgHig but a mixture of various alkanes. The 'octane rating' of petrol 
refers to the percentage of the petrol which is CsHig. For example, 93 octane fuel contains 93% CgHig 
and 7% other alkanes. The isomer of CgHig referred to in the 'octane rating' is in fact not octane but 
2,2,4-trimethylpentane. 

a. Write an unbalanced equation for the chemical reaction which takes place when petrol (CgHis) 
burns in excess oxygen. 

b. Write the general formula of the alkanes. 

c. Define the term structural isomer. 

d. Use the information given in this question and your knowledge of naming organic compounds to 
deduce and draw the full structural formula for 2,2,4-trimethylpentane. (lEB pg 25) 

1.2.1.5 The alkenes 

In the alkenes, there is at least one double bond between two carbon atoms. This means that they are 
unsaturated and are more reactive than the alkanes. The simplest alkene is ethene (also known as ethylene), 
which is shown in Figure 1.39. 



Image notjinished 



Figure 1.39: The (a) structural, (b) condensed structural and (c) molecular structure representations 
of ethene 



As with the alkanes, the alkenes also form a homologous series. They have the general formula C„H2„. 
The second alkene in the series would therefore be CsHg. This molecule is known as propene (Figure 1.40). 
Note that if an alkene has two double bonds, it is called a diene and if it has three double bonds it is called 
a triene. 



15 



Image notjinished 



Figure 1.40: The (a) structural, (b) condensed structural and (c) molecular structure representations 
of propene 



The alkenes have a variety of uses. Ethylene for example is a hormone in plants that stimulates the 
ripening of fruits and the opening of flowers. Propene is an important compound in the petrochemicals 
industry. It is used as a monomer to make polypropylene and is also used as a fuel gas for other industrial 
processes. 

1.2.1.6 Naming the alkenes 

Similar rules will apply in naming the alkenes, as for the alkanes. 

Khan academy video on alkenes 

This media object is a Flash object. Please view or download it at 
<http://www.youtube.com/v/KWv5PaoHwPA&rel=0&hl=en_US&feature=player_embedded&version=3> 



Figure 1.41 



Exercise 1.5: Naming the alkenes 

Give the lUPAC name for the following compound: 



(Solution on p. 30.) 



Image notjinished 



Figure 1.42 



Exercise 1.6: Naming the alkenes 

Draw the structural formula for the organic compound 3-methyl-butene 

Exercise 1.7: Naming the alkenes 

Give the lUPAC name for the following compound: 



(Solution on p. 30.) 



(Solution on p. 30.) 



Image notjinished 



Figure 1.43 



16 CHAPTER 1. ORGANIC MOLECULES 



1.2.1.6.1 Naming the alkenes 

Give the lUPAC name for each of the following alkenes: 



1. CH2CHCH2CH2CH3 

2. CH3CHCHCH3 



Image notjtnished 



3. 

Figure 1.44 



1.2.1.7 The properties of the alkenes 

The properties of the alkenes are very similar to those of the alkanes, except that the alkenes are more 
reactive because they are unsaturated. As with the alkanes, compounds that have four or less carbon atoms 
are gases at room temperature, while those with five or more carbon atoms are liquids. 

1.2.1.8 Reactions of the alkenes 

Alkenes can undergo addition reactions because they are unsaturated. They readily react with hydrogen, 
water and the halogens. The double bond is broken and a single, saturated bond is formed. A new group 
is then added to one or both of the carbon atoms that previously made up the double bond. The following 
are some examples: 

1. A catalyst such as platinum is normally needed for these reactions H2C = CH2 + H2 ^ H3C — CH^ 
(Figure 1.45) 



Image notjtnished 

Figure 1.45: A hydrogenation reaction 

2. CH2 = CH2 + HBr -^ CH3 - CH2 - Br (Figure 1.46) 

Image notjtnished 

Figure 1.46: A halogenation reaction 

3. CH2 = CH2 + H2O ^ CH3 - CH2 - OH (Figure 1.47) 



17 



Image notjinished 

Figure 1.47: The formation of an alcohol 



1.2.1.8.1 The Alkenes 

1. Give the lUPAC name for each of the following organic compounds: 



a. 



Image notjinished 

Figure 1.48 



b. CH3CHCH2 

2. Refer to the data table below which shows the melting point and boiling point for a number of different 
organic compounds. 



Formula 


Name 


Melting point (°C) 


Boiling point (°C) 


C4H10 


Butane 


-138 


-0.5 


C5H12 


Pentane 


-130 


36 


C6H14 


Hexane 


-95 


69 


C4H8 


Butene 


-185 


-6 


C5H10 


Pentene 


-138 


30 


C6H12 


Hexene 


-140 


63 



Table 1.5 

a. At room temperature (approx. 25°C), which of the organic compounds in the table are: 

1. gases 

2. liquids 

b. In the alkanes... 

1. Describe what happens to the melting point and boiling point as the number of carbon atoms 
in the compound increases. 

2. Explain why this is the case. 

c. If you look at an alkane and an alkene that have the same number of carbon atoms... 

1. How do their melting points and boiling points compare? 

2. Can you explain why their melting points and boiling points are different? 

d. Which of the compounds, hexane or hexene, is more reactive? Explain your answer. 

3. The following reaction takes place: CH3CHCH2 + H2 ^ CH3CH2CH3 

a. Give the name of the organic compound in the reactants. 

b. What is the name of the product? 

c. What type of reaction is this? 

d. Which compound in the reaction is a saturated hydrocarbon? 



18 CHAPTER 1. ORGANIC MOLECULES 

1.2.1.9 The Alkynes 

In the alkynes, there is at least one triple bond between two of the carbon atoms. They are unsaturated 
compounds and are therefore highly reactive. Their general formula is C„H2„_2- The simplest alkyne is 
ethyne (Figure 1.49), also known as acetylene. Many of the alkynes are used to synthesise other chemical 
products. 

Image notjinished 

Figure 1.49: Ethyne (acetylene) 



NOTE: The raw materials that are needed to make acetylene are calcium carbonate and coal. 
Acetylene can be produced through the following reactions: 

CaCOa -^ CaO 

CaO + 3C ^ CaC2 + CO 

CaC2 + 2H2O -^ Ca{OH)^ + C2H2 

An important use of acetylene is in oxyacetylene gas welding. The fuel gas burns with oxygen in a 
torch. An incredibly high heat is produced, and this is enough to melt metal. 

1.2.1.10 Naming the alkynes 

The same rules will apply as for the alkanes and alkenes, except that the sufRx of the name will now be -yne. 

Exercise 1.8: Naming the alkynes (Solution on p. 31.) 

Give the lUPAC name for the following compound: 

CKi-CH-CH,- C = C -CHt 



CH 



Figure 1.50 



1.2.1.10.1 The alkynes 

Give the lUPAC name for each of the following organic compounds. 



19 



Image notjinished 



1. 

Figure 1.51 



2. C2H2 

3. CH3CH2CCH 



1.3 The alcohols' 
1.3.1 The Alcohols 

An alcohol is any organic compound where there is a hydroxyl functional group (-0H) bound to a carbon 
atom. The general formula for a simple alcohol is C„H2„+iOH. 

The simplest and most commonly used alcohols are methanol and ethanol (Figure 1.52). 

Image notjinished 

Figure 1.52: (a) methanol and (b) ethanol 

The alcohols have a number of different uses: 

• methylated spirits (surgical spirits) is a form of ethanol where methanol has been added 

• ethanol is used in alcoholic drinks 

• ethanol is used as an industrial solvent 

• methanol and ethanol can both be used as a fuel and they burn more cleanly than gasoline or diesel 
(refer to Grade 11 for more information on biofuels as an alternative energy resource.) 

• ethanol is used as a solvent in medical drugs, perfumes and vegetable essences 

• ethanol is an antiseptic 

NOTE: Fermentation' refers to the conversion of sugar to alcohol using yeast (a fungus). The 
process of fermentation produces items such as wine, beer and yoghurt. To make wine, grape juice 
is fermented to produce alcohol. This reaction is shown below: 

CeHi20e -^ 2CO2 + 2C2H5OH + energy 

NOTE: Ethanol is a diuretic. In humans, ethanol reduces the secretion of a hormone called 
antidiuretic hormone (ADH). The role of ADH is to control the amount of water that the body 
retains. When this hormone is not secreted in the right quantities, it can cause dehyration because 
too much water is lost from the body in the urine. This is why people who drink too much alcohol 
can become dehydrated, and experience symptoms such as headaches, dry mouth, and lethargy. 
Part of the reason for the headaches is that dehydration causes the brain to shrink away from the 
skull slightly. The effects of drinking too much alcohol can be reduced by drinking lots of water. 



^This content is available online at <http://siyavula.cnx.Org/content/m39455/l.l/>. 



20 CHAPTER 1. ORGANIC MOLECULES 

1.3.1.1 Naming the alcohols 

The rules used to name the alcohols are similar to those already discussed for the other compounds, except 
that the sufRx of the name will be different because the compound is an alcohol. 

Khan academy video on alcohols 

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<http://www.youtube.com/v/nQ7QSV4JRSs&rel=0&hl=en_US&feature=player_embedded&version=3> 



Figure 1.53 



Exercise 1.9: Naming alcohols 1 (Solution on p. 31.) 

Give the lUPAC name for the following organic compound 



Image notjinished 

Figure 1.54 



Exercise 1.10: Naming alcohols 2 (Solution on p. 31.) 

Give the lUPAC name for the following compound: 



Image notjinished 

Figure 1.55 



1.3.1.1.1 Naming the alcohols 

1. Give the structural formula of each of the following organic compounds: 

a. pentan-3-ol 

b. butan-2,3-diol 

c. 2-methyl-propanol 

2. Give the lUPAC name for each of the following: 

a. CH3CH2CH(OH)CH3 



Im.age notjinished 



b. 

Figure 1.56 



21 

1.3.1.2 Physical and chemical properties of the alcohols 

The hydroxyl group affects the solubility of the alcohols. The hydroxyl group generally makes the alcohol 
molecule polar and therefore more likely to be soluble in water. However, the carbon chain resists solubility, 
so there are two opposing trends in the alcohols. Alcohols with shorter carbon chains are usually more soluble 
than those with longer carbon chains. 

Alcohols tend to have higher boiling points than the hydrocarbons because of the strong hydrogen 
bond between hydrogen and oxygen atoms. 

Alcohols can show either acidic or basic properties because of the hydroxyl group. They also undergo 
oxidation reactions to form aldehydes, ketones and carboxylic acids. 

1.3.1.2.1 Case Study : The uses of the alcohols 

Read the extract below and then answer the questions that follow: 

The alcohols are a very important group of organic compounds, and they have a variety of uses. Our 
most common use of the word 'alcohol' is with reference to alcoholic drinks. The alcohol in drinks is in 
fact ethanol. But ethanol has many more uses apart from alcoholic drinks! When ethanol burns in air, it 
produces carbon dioxide, water and energy and can therefore be used as a fuel on its own, or in mixtures with 
petrol. Because ethanol can be produced through fermentation, this is a useful way for countries without an 
oil industry to reduce imports of petrol. Ethanol is also used as a solvent in many perfumes and cosmetics. 

Methanol can also be used as a fuel, or as a petrol additive to improve combustion. Most methanol is 
used as an industrial feedstock, in other words it is used to make other things such as methanal (formalde- 
hyde), ethanoic acid and methyl esters. In most cases, these are then turned into other products. 

Propan-2-ol is another important alcohol, which is used in a variety of applications as a solvent. 

Questions 

1. Give the structural formula for propan-2-ol. 

2. Write a balanced chemical equation for the combustion reaction of ethanol. 

3. Explain why the alcohols are good solvents. 



1.4 Carboxylic acids, amines and carboxlyic acid derivatives* 

1.4,1 Carboxylic Acids 

Carboxylic acids are organic acids that are characterised by having a carboxyl group, which has the formula 
-(C=0)-OH, or more commonly written as -COOH. In a carboxyl group, an oxygen atom is double-bonded 
to a carbon atom, which is also bonded to a hydroxyl group. The simplest carboxylic acid, methanoic acid, 
is shown in Figure 1.57. The lUPAC sufRx for carboxylic acids is -anoic acid. 



Image notjinished 

Figure 1.57: Methanoic acid 



Carboxylic acids are widespread in nature. Methanoic acid (also known as formic acid) has the formula 
HCOOH and is found in insect stings. Ethanoic acid (CH3COOH), or acetic acid, is the main component of 



*This content is available online at <http://siyavula.cnx.Org/content/m39457/l.l/>. 



22 



CHAPTER 1. ORGANIC MOLECULES 



vinegar. More complex organic acids also have a variety of different functions. Benzoic acid (CeHsCOOH) 
for example, is used as a food preservative. 

NOTE: A certain type of ant, called formicine ants, manufacture and secrete formic acid, which is 
used to defend themselves against other organisms that might try to eat them. 



1.4.1.1 Physical Properties 

Carboxylic acids are weak acids, in other words they only dissociate partially. Why does the carboxyl 
group have acidic properties? In the carboxyl group, the hydrogen tends to separate itself from the oxygen 
atom. In other words, the carboxyl group becomes a source of positively-charged hydrogen ions (H+). This 
is shown in Figure 1.58. 



Image notjtnished 



Figure 1.58: The dissociation of a carboxylic acid 



1.4.1.1.1 Carboxylic acids 

1. Refer to the table below which gives information about a number of carboxylic acids, and then answer 
the questions that follow. 



Formula 


Common 
name 


Source 


lUPAC 

name 


melting 
point (OC) 


boiling 
point (OC) 




formic acid 


ants 


methanoic 
acid 


8.4 


101 


CH3CO2H 




vinegar 


ethanoic acid 


16.6 


118 




propionic acid 


milk 


propanoic 
acid 


-20.8 


141 


CH3(CH2)2CO; 


;H)utyric acid 


butter 




-5.5 


164 




valeric acid 


valerian root 


pentanoic 
acid 


-34.5 


186 


CH3(CH2)4CO; 


iHaproic acid 


goats 




-4 


205 




enanthic acid 


vines 




-7.5 


223 


continued on next page 



23 



CH3(CH2)6CO; 


iHaprylic acid 


goats 




16.3 


239 



Table 1.6 

a. Fill in the missing spaces in the table by writing the formula, common name or lUPAC name. 

b. Draw the structural formula for butyric acid. 

c. Give the molecular formula for caprylic acid. 

d. Draw a graph to show the relationship between molecular mass (on the x-axis) and boiling point 
(on the y-axis) 

1. Describe the trend you see. 

2. Suggest a reason for this trend. 

1.4.1.2 Derivatives of carboxylic acids: The esters 

When an alcohol reacts with a carboxylic acid, an ester is formed. Most esters have a characteristic and 
pleasant smell. In the reaction, the hydrogen atom from the hydroxyl group, and an OH from the carboxlic 
acid, form a molecule of water. A new bond is formed between what remains of the alcohol and acid. The 
name of the ester is a combination of the names of the alcohol and carboxylic acid. The sufRx for an ester 
is -oate. An example is shown in Figure 1.59. 

Image notjinished 

Figure 1.59: The formation of an ester and water from an alcohol and carboxylic acid 



1,4,2 The Amino Group 

The amino group has the formula -NH2 and consists of a nitrogen atom that is bonded to two hydrogen 
atoms, and to the carbon skeleton. Organic compounds that contain this functional group are called amines. 
One example is glycine. Glycine belongs to a group of organic compounds called amino acids, which are the 
building blocks of proteins. 

Image notjinished 

Figure 1.60: A molecule of glycine 



1,4,3 The Carbonyl Group 

The carbonyl group (-CO) consists of a carbon atom that is joined to an oxygen by a double bond. If the 
functional group is on the end of the carbon chain, the organic compound is called a ketone. The simplest 
ketone is acetone, which contains three carbon atoms. A ketone has the ending 'one' in its lUPAC name. 



24 



CHAPTER 1. ORGANIC MOLECULES 



1.4.3.1 Carboxylic acids, esters, amines and ketones 

1. Look at the list of organic compounds in the table below: 



Organic compound 


Type of compound 


CH3CH2CH2COOH 




NH2CH2COOH 




propyl ethanoate 




CH3CHO 





Table 1.7 

a. Complete the table by identifying each compound as either a carboxylic acid, ester, amine or 
ketone. 

b. Give the name of the compounds that have been written as condensed structural formulae. 

2. A chemical reaction takes place and ethyl methanoate is formed. 

a. What type of organic compound is ethyl methanoate? 

b. Name the two reactants in this chemical reaction. 

c. Give the structural formula of ethyl methanoate. 

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Figure 1.61 



1.4.4 Summary 

• Organic chemistry is the branch of chemistry that deals with organic molecules. An organic 
molecule is one that contains carbon. 

• All living organisms contain carbon. Plants use sunlight to convert carbon dioxide in the air into 
organic compounds through the process of photosynthesis. Animals and other organisms then feed 
on plants to obtain their own organic compounds. Fossil fuels are another important source of carbon. 

• It is the unique properties of the carbon atom that give organic compounds certain properties. 

• The carbon atom has four valence electrons, so it can bond with many other atoms, often resulting 
in long chain structures. It also forms mostly covalent bonds with the atoms that it bonds to, 
meaning that most organic molecules are non-polar. 

• An organic compound can be represented in different ways, using its molecular formula, structural 
formula or condensed structural formula. 

• If two compounds are isomers, it means that they have the same molecular formulae but different 
structural formulae. 

• A functional group is a particular group of atoms within a molecule, which give it certain reaction 
characteristics. Organic compounds can be grouped according to their functional group. 



25 

The hydrocarbons are organic compounds that contain only carbon and hydrogen. They can be 

further divided into the alkanes, alkenes and alkynes, based on the type of bonds between the carbon 

atoms. 

The alkanes have only single bonds between their carbon atoms and are unreactive. 

The alkenes have at least one double bond between two of their carbon atoms. They are more 

reactive than the alkanes. 

The alkynes have at least one triple bond between two of their carbon atoms. They are the most 

reactive of the three groups. 

A hydrocarbon is said to be saturated if it contains the maximum possible number of hydrogen atoms 

for that molecule. The alkanes are all saturated compounds. 

A hydrocarbon is unsaturated if it does not contain the maximum number of hydrogen atoms for 

that molecule. The alkenes and alkynes are examples of unsaturated molecules. If a double or triple 

bond is broken, more hydrogen atoms can be added to the molecule. 

There are three types of reactions that occur in the alkanes: substitution, elimination and oxidation 

reactions. 

The alkenes undergo addition reactions because they are unsaturated. 

Organic compounds are named according to their functional group and its position in the molecule, 

the number of carbon atoms in the molecule and the position of any double and triple bonds. The 

lUPAC rules for nomenclature are used in the naming of organic molecules. 

Many of the properties of the hydrocarbons are determined by their molecular structure, the 

bonds between atoms and molecules, and their surface area. 

The melting point and boiling point of the hydrocarbons increases as their number of carbon atoms 

increases. 

The molecular mass of the hydrocarbons determines whether they will be in the gaseous, liquid or 

solid phase at certain temperatures. 

An alcohol is an organic compound that contains a hydroxyl group (-0H). 

The alcohols have a number of different uses including their use as a solvent, for medicinal purposes 

and in alcoholic drinks. 

The alcohols share a number of properties because of the hydroxyl group. The hydroxyl group affects 

the solubility of the alcohols. Those with shorter carbon chains are generally more soluble, and those 

with longer chains are less soluble. The strong hydrogen bond between the hydrogen and oxygen 

atoms in the hydroxyl group gives alcohols a higher melting point and boiling point than other organic 

compounds. The hydroxyl group also gives the alcohols both acidic and basic properties. 

The carboxylic acids are organic acids that contain a carboxyl group with the formula -COOH. 

In a carboxyl group, an oxygen atom is double-bonded to a carbon atom, which is also bonded to a 

hydroxyl group. 

The carboxylic acids have weak acidic properties because the hydrogen atom is able to dissociate 

from the carboxyl group. 

An ester is formed when an alcohol reacts with a carboxylic acid. 

The amines are organic compounds that contain an amino functional group, which has the formula 

-NH2. Some amines belong to the amino acid group, which are the building blocks of proteins. 

The ketones are a group of compounds that contain a carbonyl group, which consists of an oxygen 

atom that is double-bonded to a carbon atom. In a ketone, the carbonyl group is on the end of the 

carbon chain. 



1.4.4.1 Summary exercise 

1. Give one word for each of the following descriptions: 

a. The group of hydrocarbons to which 2-methyl-propene belongs. 

b. The name of the functional group that gives alcohols their properties. 

c. The group of organic compounds that have acidic properties. 



26 



CHAPTER 1. ORGANIC MOLECULES 



d. The name of the organic compound that is found in vinegar. 

e. The name of the organic compound that is found in alcoholic beverages. 

2. In each of the following questions, choose the one correct answer from the list provided. 

a. When 1-propanol is oxidised by acidified potassium permanganate, the possible product formed 
is... 

1. propane 

2. propanoic acid 

3. methyl propanol 

4. propyl methanoate 
(lEB 2004) 

b. What is the lUPAC name for the compound represented by the following structural formula? 



Image notjtnished 



Figure 1.62 



1. 1,2,2-trichlorobutane 

2. l-chloro-2,2-dichlorobutane 

3 . 1 , 2 , 2-t richloro-3-met hy Ipropane 

4. l-chloro-2,2-dichloro-3-methylpropane 

(lEB 2003) 

3. Write balanced equations for the following reactions: 

a. Ethene reacts with bromine 

b. Ethyne gas burns in an excess of oxygen 

c. Ethanoic acid ionises in water 

4. The table below gives the boiling point of ten organic compounds. 





Compound 


Formula 


Boiling Point (°C) 


1 


methane 


CH4 


-164 


2 


ethane 


C2H6 


-88 


3 


propane 


C3H8 


-42 


4 


butane 


C4H10 





5 


pentane 


C5H12 


36 


6 


methanol 


CH3OH 


65 


7 


ethanol 


C2H5OH 


78 


8 


propan-1-ol 


C3H7OH 


98 


9 


propan-l,2-diol 


CH3CHOHCH2OH 


189 


10 


propan-l,2,3-triol 


CH2OHCHOHCH2OH 


290 



Table 1.8 

The following questions refer to the compounds shown in the above table. 



27 

a. To which homologous series do the following compounds belong? 

1. Compounds 1,2 and 3 

2. Compounds 6,7 and 8 

b. Which of the above compounds are gases at room temperature? 

c. What causes the trend of increasing boiling points of compounds 1 to 5? 

d. Despite the fact that the length of the carbon chain in compounds 8,9 and 10 is the same, the 
boiling point of propan-l,2,3-triol is much higher than the boiling point of propan-1-ol. What is 
responsible for this large difference in boiling point? 

e. Give the lUPAC name and the structural formula of an isomer of butane. 

f. Which one of the above substances is used as a reactant in the preparation of the ester ethyl- 
methanoate? 

g. Using structural formulae, write an equation for the reaction which produces ethylmethanoate. 

{lEB 2004) 
5. Refer to the numbered diagrams below and then answer the questions that follow. 



Image notjinished 

Figure 1.63 



a. Which one of the above compounds is produced from the fermentation of starches and sugars in 
plant matter? 

1. compound 1 

2. compound 2 

3. compound 3 

4. compound 4 

b. To which one of the following homologous series does compound 1 belong? 

1. esters 

2. alcohols 

3. aldehydes 

4. carboxylic acids 

c. The correct lUPAC name for compound 3 is... 

1. l,l-dibromo-3-butyne 

2. 4,4-dibromo-l-butyne 

3. 2,4-dibromo-l-butyne 

4. 4,4-dibromo-l-propyne 

d. What is the correct lUPAC name for compound 4? 

1. propanoic acid 

2. ethylmethanoate 

3. methylethanoate 

4. methylpropanoate 

lEB 2005 
6. Answer the following questions: 

a. What is a homologous series? 

b. A mixture of ethanoic acid and methanol is warmed in the presence of concentrated sulphuric 
acid. 



28 CHAPTER 1. ORGANIC MOLECULES 

1. Using structural formulae, give an equation for the reaction which takes place. 

2. What is the lUPAC name of the organic compound formed in this reaction? 
c. Consider the following unsaturated hydrocarbon: 

Image notjtnished 

Figure 1.64 



1. Give the lUPAC name for this compound. 

2. Give the balanced equation for the combustion of this compound in excess oxygen. 

(lEB Paper 2, 2003) 
7. Consider the organic compounds labelled A to E. A. CH3CH2CH2CH2CH2CH3 B. CgHe C. CH3-CI 
D. Methylamine 



Image notjtnished 

Figure 1.65 



a. Write a balanced chemical equation for the preparation of compound C using an alkane as one of 
the reactants. 

b. Write down the lUPAC name for compound E. 

c. Write down the structural formula of an isomer of compound A that has only FOUR carbon atoms 
in the longest chain. 

d. Write down the structural formula for compound B. 



29 

Solutions to Exercises in Chapter 1 

Solution to Exercise 1.1 (p. 10) 

Step 1. The compound is a hydrocarbon with single bonds between the carbon atoms. It is an alkane and will 

have a sufRx of -ane. 
Step 2. There are four carbon atoms in the longest chain. The prefix of the compound will be 'but'. 
Step 3. In this case, it is easy. The carbons are numbered from left to right, from one to four. 
Step 4. There are no branched groups in this compound. 
Step 5. The name of the compound is butane. 

Solution to Exercise 1.2 (p. 11) 

Step 1. The compound is an alkane and will have the sufRx -ane. 

Step 2. There are three carbons in the longest chain. The prefix for this compound is -prop. 

Step 3. If we start at the carbon on the left, we can number the atoms as shown below: 



Image notjinished 

Figure 1.66 



Step 4. There is a branched group attached to the second carbon atom. This group has the formula CH3 which 
is methane. However, because it is not part of the main chain, it is given the sufRx -yl (i.e. methyl). 
The position of the methyl group comes just before its name (see next step). 

Step 5. The compound's name is 2-methylpropane. 

Solution to Exercise 1.3 (p. 11) 

Step 1. The structural formula of the compound is: 



Image notjinished 

Figure 1.67 



Step 2. The compound is an alkane and will have the sufRx -ane. 

Step 3. There are four carbons in the longest chain. The preRx for this compound is -but. 
Step 4. If we start at the carbon on the left, carbon atoms are numbered as shown in the diagram above. A 
second way that the carbons could be numbered is: 



Image notjinished 

Figure 1.68 



30 CHAPTER 1. ORGANIC MOLECULES 

Step 5. There are two methyl groups attached to the main chain. The first one is attached to the second 
carbon atom and the second methyl group is attached to the third carbon atom. Notice that in this 
example it does not matter how you have chosen to number the carbons in the main chain; the methyl 
groups are still attached to the second and third carbons and so the naming of the compound is not 
affected. 

Step 6. The compound's name is 2,3-dimethyl-butane. 

Solution to Exercise 1.4 (p. 11) 

Step 1. The compound is an alkane and will have the sufRx -ane. 

Step 2. There are six carbons in the longest chain if they are numbered as shown below. The prefix for the 
compound is hex-. 



Image not finished 

Figure 1.69 



Step 3. There is one methyl group attached to the main chain. This is attached to the third carbon atom. 
Step 4. The compound's name is 3-methyl-hexane. 

Solution to Exercise 1.5 (p. 15) 

Step 1. The compound is an alkene and will have the sufRx -ene. 

Step 2. There are four carbon atoms in the longest chain and so the prefix for this compound will be 'but'. 

Step 3. Remember that when there is a double or triple bond, the carbon atoms must be numbered so that 
the double or triple bond is at the lowest numbered carbon. In this case, it doesn't matter whether 
we number the carbons from the left to right, or from the right to left. The double bond will still fall 
between C2 and C3. The position of the bond will come just before the sufRx in the compound's name. 

Step 4. There are no branched groups in this molecule. 

Step 5. The name of this compound is but-2-ene. 

Solution to Exercise 1.6 (p. 15) 

Step 1. The sufRx -ene means that this compound is an alkene and there must be a double bond in the molecule. 

There is no number immediately before the sufRx which means that the double bond must be at the 

Rrst carbon in the chain. 
Step 2. The preRx for the compound is 'but' so there must be four carbons in the longest chain. 
Step 3. There is a methyl group at the third carbon atom in the chain. 



Image notjinished 



Step 4. 

Figure 1.70 



Solution to Exercise 1.7 (p. 15) 

Step 1. The compound is an alkene and will have the sufRx -ene. There is a double bond between the Rrst and 
second carbons and also between the third and forth carbons. The organic compound is therefore a 
'diene'. 



31 

Step 2. There are four carbon atoms in the longest chain and so the prefix for this compound will be 'but'. 

The carbon atoms are numbered 1 to 4 in the diagram above. Remember that the main carbon chain 

must contain both the double bonds. 
Step 3. There is an ethyl group on the second carbon. 
Step 4. The name of this compound is 2-ethyl-but-l,3-diene. 

Solution to Exercise 1.8 (p. 18) 

Step 1. There is a triple bond between two of the carbon atoms, so this compound is an alkyne. The suffix 

will be -yne. The triple bond is at the second carbon, so the suffix will in fact be 2-yne. 
Step 2. If we count the carbons in a straight line, there are six. The prefix of the compound's name will be 

'hex'. 
Step 3. In this example, you will need to number the carbons from right to left so that the triple bond is 

between carbon atoms with the lowest numbers. 
Step 4. There is a methyl (CH3) group attached to the fifth carbon (remember we have numbered the carbon 

atoms from right to left). 
Step 5. If we follow this order, the name of the compound is 5-methyl-hex-2-yne. 

Solution to Exercise 1.9 (p. 20) 

Step 1. The compound has an -OH (hydroxyl) functional group and is therefore an alcohol. The compound 

will have the suffix -ol. 
Step 2. There are three carbons in the longest chain. The prefix for this compound will be 'prop'. Since there 

are only single bonds between the carbon atoms, the suffix becomes 'propan' (similar to the alkane 

'propane'). 
Step 3. In this case, it doesn't matter whether you start numbering from the left or right. The hydroxyl group 

will still be attached to the middle carbon atom, numbered '2'. 
Step 4. There are no branched groups in this compound, but you still need to indicate the position of the 

hydroxyl group on the second carbon. The suffix will be -2-ol because the hydroxyl group is attached 

to the second carbon. 
Step 5. The compound's name is propan-2-ol. 

Solution to Exercise 1.10 (p. 20) 

Step 1. The compound has an -OH (hydroxyl) functional group and is therefore an alcohol. There are two 

hydroxyl groups in the compound, so the suffix will be -diol. 
Step 2. There are four carbons in the longest chain. The prefix for this compound will be 'butan'. 
Step 3. The carbons will be numbered from left to right so that the two hydroxyl groups are attached to carbon 

atoms with the lowest numbers. 
Step 4. There are no branched groups in this compound, but you still need to indicate the position of the 

hydroxyl groups on the first and second carbon atoms. The suffix will therefore become 1,2-diol. 
Step 5. The compound's name is butan- 1,2-diol. 



32 CHAPTER 1. ORGANIC MOLECULES 



Chapter 2 

Organic macromolecules 

2.1 Polymers' 

2.1.1 Organic Macromolecules 

As its name suggests, a macromolecule is a large molecule that forms when lots of smaller molecules are 
joined together. In this chapter, we will be taking a closer look at the structure and properties of different 
macromolecules, and at how they form. 

2.1.2 Polymers 

Some macromolecules are made up of lots of repeating structural units called monomers. To put it more 
simply, a monomer is like a building block. When lots of similar monomers are joined together by covalent 
bonds, they form a polymer. In an organic polymer, the monomers are joined by the carbon atoms of 
the polymer 'backbone'. A polymer can also be inorganic, in which case there may be atoms such as silicon 
in the place of carbon atoms. The key feature that makes a polymer different from other macromolecules, is 
the repetition of identical or similar monomers in the polymer chain. The examples shown below will help 
to make these concepts clearer. 

Definition 2.1: Polymer 

Polymer is a term used to describe large molecules consisting of repeating structural units, or 
monomers, connected by covalent chemical bonds. 

1. Polyethene Chapter looked at the structure of a group of hydrocarbons called the alkenes. One 
example is the molecule ethene. The structural formula of ethene is is shown in Figure 2.1. When lots 
of ethene molecules bond together, a polymer called polyethene is formed. Ethene is the monomer 
which, when joined to other ethene molecules, forms the poiymerpolyethene. Polyethene is a cheap 
plastic that is used to make plastic bags and bottles. 

Image notjinished 

Figure 2.1: (a) Ethene monomer and (b) polyethene polymer 



A polymer may be a chain of thousands of monomers, and so it is impossible to draw the entire polymer. 
Rather, the structure of a polymer can be condensed and represented as shown in Figure 2.2. The 



^This content is available online at <http://siyavula.cnx.Org/content/m39435/l.l/>. 

33 



34 CHAPTER 2. ORGANIC MACROMOLECULES 

monomer is enclosed in brackets and the 'n' represents the number of ethene molecules in the polymer, 
where 'n' is any whole number. What this shows is that the ethene monomer is repeated an indefinite 
number of times in a molecule of polyethene. 



Image notjinished 

Figure 2.2: A simplified representation of a polyethene molecule 



2. Polypropene Another example of a polymer is polypropene (fig Figure 2.3). Polypropene is also a 
plastic, but is stronger than polyethene and is used to make crates, fibres and ropes. In this polymer, 
the monomer is the alkene called propene. 



Image notjinished 

Figure 2.3: (a) Propene monomer and (b) polypropene polymer 



2.1.3 How do polymers form? 

Polymers are formed through a process called polymerisation, where monomer molecules react together 
to form a polymer chain. Two types of polymerisation reactions are addition polymerisation and con- 
densation polymerisation. 

Definition 2.2: Polymerisation 

In chemistry, polymerisation is a process of bonding monomers, or single units together through a 
variety of reaction mechanisms to form longer chains called polymers. 

2.1.3.1 Addition polymerisation 

In this type of reaction, monomer molecules are added to a growing polymer chain one at a time. No small 
molecules are eliminated in the process. An example of this type of reaction is the formation of polyethene 
from ethene (fig Figure 2.1). When molecules of ethene are joined to each other, the only thing that changes 
is that the double bond between the carbon atoms in each ethene monomer is replaced by a single bond so 
that a new carbon-carbon bond can be formed with the next monomer in the chain. In other words, the 
monomer is an unsaturated compound which, after an addition reaction, becomes a saturated compound. 

2.1.3.1.1 Initiation, propagation and termination 

There are three stages in the process of addition polymerisation. Initiation refers to a chemical reaction 
that triggers off another reaction. In other words, initiation is the starting point of the polymerisation 
reaction. Chain propagation is the part where monomers are continually added to form a longer and 
longer polymer chain. During chain propagation, it is the reactive end groups of the polymer chain that 
react in each propagation step, to add a new monomer to the chain. Once a monomer has been added, the 
reactive part of the polymer is now in this last monomer unit so that propagation will continue. Termination 
refers to a chemical reaction that destroys the reactive part of the polymer chain so that propagation stops. 

Exercise 2.1: Polymerisation reactions (Solution on p. 52.) 

A polymerisation reaction takes place and the following polymer is formed: 



35 



Image notjinished 



Figure 2.4 



Note: W, X, Y and Z could represent a number of different atoms or combinations of atoms e.g. 
H, F, CI or CH3. 

1. Give the structural formula of the monomer of this polymer. 

2. To what group of organic compounds does this monomer belong? 

3. What type of polymerisation reaction has taken place to join these monomers to form the 
polymer? 



2.1.3.2 Condensation polymerisation 

In this type of reaction, two monomer molecules form a covalent bond and a small molecule such as water 
is lost in the bonding process. Nearly all biological reactions are of this type. Polyester and nylon are 
examples of polymers that form through condensation polymerisation. 

1. Polyester Polyesters are a group of polymers that contain the ester functional group in their main 
chain. Although there are many forms of polyesters, the term polyester usually refers to polyethylene 
terephthalate (PET). PET is made from ethylene glycol (an alcohol) and terephthalic acid (a carboxylic 
acid). In the reaction, a hydrogen atom is lost from the alcohol, and a hydroxyl group is lost from the 
carboxylic acid. Together these form one water molecule which is lost during condensation reactions. 
A new bond is formed between an oxygen and a carbon atom. This bond is called an ester linkage. 
The reaction is shown in Figure 2.5. 



Image notjinished 



Figure 2.5: An acid and an alcohol monomer react (a) to form a molecule of the polyester 'polyethylene 
terephthalate' (b). 



Polyesters have a number of characteristics which make them very useful. They are resistant to 
stretching and shrinking, they are easily washed and dry quickly, and they are resistant to mildew. It 
is for these reasons that polyesters are being used more and more in textiles. Polyesters are stretched 
out into fibres and can then be made into fabric and articles of clothing. In the home, polyesters are 
used to make clothing, carpets, curtains, sheets, pillows and upholstery. 

NOTE: Polyester is not just a textile. Polyethylene terephthalate is in fact a plastic which can 
also be used to make plastic drink bottles. Many drink bottles are recycled by being reheated 
and turned into polyester fibres. This type of recycling helps to reduce disposal problems. 

2. Nylon Nylon was the first polymer to be commercially successful. Nylon replaced silk, and was used 
to make parachutes during World War 2. Nylon is very strong and resistant, and is used in fishing 
line, shoes, toothbrush bristles, guitar strings and machine parts to name just a few. Nylon is formed 
from the reaction of an amine (1,6-diaminohexane) and an acid monomer (adipic acid) (Figure 2.6). 



36 CHAPTER 2. ORGANIC MACROMOLECULES 

The bond that forms between the two monomers is called an amide linkage. An amide linkage forms 
between a nitrogen atom in the amine monomer and the carbonyl group in the carboxylic acid. 

Image notjinished 

Figure 2.6: An amine and an acid monomer (a) combine to form a section of a nylon polymer (b). 



NOTE: Nylon was first introduced around 1939 and was in high demand to make stockings. 
However, as World War 2 progressed, nylon was used more and more to make parachutes, and so 
stockings became more difficult to buy. After the war, when manufacturers were able to shift their 
focus from parachutes back to stockings, a number of riots took place as women queued to get 
stockings. In one of the worst disturbances, 40 000 women queued up for 13 000 pairs of stockings, 
which led to fights breaking out! 



2.1.3.2.1 Polymers 

1. The following monomer is a reactant in a polymerisation reaction: 



Image notjinished 



Figure 2.7 



a. What is the lUPAC name of this monomer? 

b. Give the structural formula of the polymer that is formed in this polymerisation reaction. 

c. Is the reaction an addition or condensation reaction? 



2. The polymer below is the product of a polymerisation reaction. 

Image notjinished 

Figure 2.8 



a. Give the structural formula of the monomer in this polymer. 

b. What is the name of the monomer? 

c. Draw the abbreviated structural formula for the polymer. 

d. Has this polymer been formed through an addition or condensation polymerisation reaction? 

3. A condensation reaction takes place between methanol and methanoic acid, 
a. Give the structural formula for... 
1. methanol 



37 

2. methanoic acid 

3. the product of the reaction 

b. What is the name of the product? (Hint: The product is an ester) 

2.1.4 The chemical properties of polymers 

The attractive forces between polymer chains play a large part in determining a polymer's properties. Because 
polymer chains are so long, these interchain forces are very important. It is usually the side groups on the 
polymer that determine what types of intermolecular forces will exist. The greater the strength of the 
intermolecular forces, the greater will be the tensile strength and melting point of the polymer. Below are 
some examples: 

• Hydrogen bonds between adjacent chains Polymers that contain amide or carbonyl groups can 
form hydrogen bonds between adjacent chains. The positive hydrogen atoms in the N-H groups of 
one chain are strongly attracted to the oxygen atoms (more precisely, the lone-pairs on the oxygen) 
in the C=0 groups on another. Polymers that contain urea linkages would fall into this category. 
The structural formula for urea is shown in Figure 2.9. Polymers that contain urea linkages have high 
tensile strength and a high melting point. 



Image notjinished 

Figure 2.9: The structural formula for urea 



Dipole-dipole bonds between adjacent chains Polyesters have dipole-dipole bonding between 
their polymer chains. Dipole bonding is not as strong as hydrogen bonding, so a polyester's melting 
point and strength are lower than those of the polymers where there are hydrogen bonds between the 
chains. However, the weaker bonds between the chains means that polyesters have greater flexibility. 
The greater the flexibility of a polymer, the more likely it is to be moulded or stretched into fibres. 
Weak van der Waal's forces Other molecules such as ethene do not have a permanent dipole and 
so the attractive forces between polyethene chains arise from weak van der Waals forces. Polyethene 
therefore has a lower melting point than many other polymers. 



2.1.5 Types of polymers 

There are many different types of polymers. Some are organic, while others are inorganic. Organic polymers 
can be broadly grouped into either synthetic/semi-synthetic (artificial) or biological (natural) polymers. We 
are going to take a look at two groups of organic polymers: plastics, which are usually synthetic or semi- 
synthetic and biological macromolecules which are natural polymers. Both of these groups of polymers play 
a very important role in our lives. 

2.1.6 Plastics 

In today's world, we can hardly imagine life without plastic. From cellphones to food packaging, fishing line 
to plumbing pipes, compact discs to electronic equipment, plastics have become a very important part of our 
daily lives. "Plastics" cover a range of synthetic and semi-synthetic organic polymers. Their name comes 
from the fact that they are 'malleable', in other words their shape can be changed and moulded. 



38 CHAPTER 2. ORGANIC MACROMOLECULES 

Definition 2.3: Plastic 

Plastic covers a range of synthetic or semisynthetic organic polymers. Plastics may contain other 
substances to improve their performance. Their name comes from the fact that many of them are 
malleable, in other words they have the property of plasticity. 

It was only in the nineteenth century that it was discovered that plastics could be made by chemically 
changing natural polymers. For centuries before this, only natural organic polymers had been used. Examples 
of natural organic polymers include waxes from plants, cellulose (a plant polymer used in fibres and ropes) 
and natural rubber from rubber trees. But in many cases, these natural organic polymers didn't have the 
characteristics that were needed for them to be used in specific ways. Natural rubber for example, is sensitive 
to temperature and becomes sticky and smelly in hot weather and brittle in cold weather. 

In 1834 two inventors, Friedrich Ludersdorf of Germany and Nathaniel Hayward of the US, independently 
discovered that adding sulfur to raw rubber helped to stop the material from becoming sticky. After this, 
Charles Goodyear discovered that heating this modified rubber made it more resistant to abrasion, more 
elastic and much less sensitive to temperature. What these inventors had done was to improve the properties 
of a natural polymer so that it could be used in new ways. An important use of rubber now is in vehicle 
tyres, where these properties of rubber are critically important. 

NOTE: The first true plastic (i.e. one that was not based on any material found in nature) was 
Bakelite, a cheap, strong and durable plastic. Some of these plastics are still used for example in 
electronic circuit boards, where their properties of insulation and heat resistance are very important. 



2.1.6.1 The uses of plastics 

There is such a variety of different plastics available, each having their own specific properties and uses. The 
following are just a few examples. 

• Polystyrene Polystyrene (Figure 2.10) is a common plastic that is used in model kits, disposable 
eating utensils and a variety of other products. In the polystyrene polymer, the monomer is styrene, 
a liquid hydrocarbon that is manufactured from petroleum. 

Image notjinished 

Figure 2.10: The polymerisation of a styrene monomer to form a polystyrene polymer 

• Polyvinylchloride (PVC) Polyvinyl chloride (PVC) (Figure 2.11) is used in plumbing, gutters, 
electronic equipment, wires and food packaging. The side chains of PVC contain chlorine atoms, which 
give it its particular characteristics. 

Image notjinished 

Figure 2.11: Polyvinyl chloride 



NOTE: Many vinyl products have other chemicals added to them to give them particular 
properties. Some of these chemicals, called additives, can leach out of the vinyl products. In 
PVC, plasticizers are used to make PVC more flexible. Because many baby toys are made 



39 



from PVC, there is concern that some of these products may leach into the mouths of the 
babies that are chewing on them. In the USA, most companies have stopped making PVC 
toys. There are also concerns that some of the plasticizers added to PVC may cause a number 
of health conditions including cancer. 

Synthetic rubber Another plastic that was critical to the World War 2 effort was synthetic rubber, 
which was produced in a variety of forms. Not only were worldwide natural rubber supplies limited, 
but most rubber-producing areas were under Japanese control. Rubber was needed for tyres and parts 
of war machinery. After the war, synthetic rubber also played an important part in the space race and 
nuclear arms race. 

Polyethene/polyethylene (PE) Polyethylene (Figure 2.1) was discovered in 1933. It is a cheap, 
fiexible and durable plastic and is used to make films and packaging materials, containers and car 
fittings. One of the most well known polyethylene products is 'Tupperware', the sealable food containers 
designed by Earl Tupper and promoted through a network of housewives! 

Polytetrafluoroethylene (PTFE) Polytetrafiuoroethylene (Figure 2.12) is more commonly known 
as 'Tefion' and is most well known for its use in non-stick frying pans. Tefion is also used to make the 
breathable fabric Gore- Tex. 



Image notjinished 



Figure 2.12: A tetra fluoroethylene monomer and polytetrafluoroethylene polymer 



Table 2.1 summarises the formulae, properties and uses of some of the most common plastics. 



Name 


Formula 


Monomer 


Properties 


Uses 


Polyethene (low 
density) 


-(CH2-CH2)„- 


CH2=CH2 


soft, waxy soHd 


film wrap and 
plastic bags 


Polyethene (high 
density) 


-(CH2-CH2)„- 


CH2=CH2 


rigid 


electrical insula- 
tion, bottles and 
toys 


Polypropene 


-[CH2- 
CH(CH3)]„- 


0x12^^x10x13 


different grades: 
some are soft and 
others hard 


carpets and uphol- 
stery 


Polyvinylchloride 
(PVC) 


-(CH2-CHC1)„- 


CH2=CHC1 


strong, rigid 


pipes, fiooring 


Polystyrene 


-[CH2- 
CH(C6H5)]„ 


CH2=CHC6H5 


hard, rigid 


toys, packaging 


continued on next page 



40 



CHAPTER 2. ORGANIC MACROMOLECULES 



PolytetrafiuoroethyleH^F2-CF2)„ 



CF2— CF2 



resistant, smooth, 
solid 



non-stick sur- 

faces, electrical 
insulation 



Table 2.1: A summary of the formulae, properties and uses of some common plastics 



2.1.6.1.1 Plastics 

1. It is possible for macromolecules to be composed of more than one type of repeating monomer. The 
resulting polymer is called a copolymer. Varying the monomers that combine to form a polymer, is 
one way of controlling the properties of the resulting material. Refer to the table below which shows 
a number of different copolymers of rubber, and answer the questions that follow: 



Monomer A 


Monomer B 


Copolymer 


Uses 


H2C=CHC1 


H2C=CCl2 


Saran 


films and fibres 


H2C=CHC6H5 


H2C=C-CH=CH2 


SBR (styrene butadiene rubber) 


tyres 


H2C=CHCN 


rl2vj^ 0-0x1^0x12 


Nitrile rubber 


adhesives and hoses 


H2C = C(CH3)2 


xl2C^C-Cxl^Cxl2 


Butyl rubber 


inner tubes 


F2C=CF(CF3) 


H2C=CHF 


Viton 


gaskets 



Table 2.2 

a. Give the structural formula for each of the monomers of nitrile rubber. 

b. Give the structural formula of the copolymer viton. 

c. In what ways would you expect the properties of SBR to be different from nitrile rubber? 

d. Suggest a reason why the properties of these polymers are different. 

2. In your home, find as many examples of different types of plastics that you can. Bring them to 
school and show them to your group. Together, use your examples to complete the following ta- 
ble: 



Object 


Type of plastic 


Properties 


Uses 











































Table 2.3 



2.1.6.2 Thermoplastics and thermosetting plastics 

A thermoplastic is a plastic that can be melted to a liquid when it is heated and freezes to a brittle, 
glassy state when it is cooled enough. These properties of thermoplastics are mostly due to the fact that the 
forces between chains are weak. This also means that these plastics can be easily stretched or moulded into 



41 

any shape. Examples of thermoplastics include nylon, polystyrene, polyethylene, polypropylene and PVC. 
Thermoplastics are more easily recyclable than some other plastics. 

Thermosetting plastics differ from thermoplastics because once they have been formed, they cannot 
be remelted or remoulded. Examples include bakelite, vulcanised rubber, melanine (used to make furniture), 
and many glues. Thermosetting plastics are generally stronger than thermoplastics and are better suited to 
being used in situations where there are high temperatures. They are not able to be recycled. Thermosetting 
plastics have strong covalent bonds between chains and this makes them very strong. 

2.1.6.2.1 Case Study : Biodegradable plastics 

Read the article below and then answer the questions that follow. 

Our whole world seems to be wrapped in plastic. Almost every product we buy, most of the food we eat 
and many of the liquids we drink come encased in plastic. Plastic packaging provides excellent protection 
for the product, it is cheap to manufacture and seems to last forever. Lasting forever, however, is proving 
to be a major environmental problem. Another problem is that traditional plastics are manufactured from 
non-renewable resources - oil, coal and natural gas. In an effort to overcome these problems, researchers and 
engineers have been trying to develop biodegradable plastics that are made from renewable resources, such 
as plants. 

The term biodegradable means that a substance can be broken down into simpler substances by the 
activities of living organisms, and therefore is unlikely to remain in the environment. The reason most 
plastics are not biodegradable is because their long polymer molecules are too large and too tightly bonded 
together to be broken apart and used by decomposer organisms. However, plastics based on natural plant 
polymers that come from wheat or corn starch have molecules that can be more easily broken down by 
microbes. 

Starch is a natural polymer. It is a white, granular carbohydrate produced by plants during photosyn- 
thesis and it serves as the plant's energy store. Many plants contain large amounts of starch. Starch can be 
processed directly into a bioplastic but, because it is soluble in water, articles made from starch will swell 
and deform when exposed to moisture, and this limits its use. This problem can be overcome by changing 
starch into a different polymer. First, starch is harvested from corn, wheat or potatoes, then microorganisms 
transform it into lactic acid, a monomer. Finally, the lactic acid is chemically treated to cause the molecules 
of lactic acid to link up into long chains or polymers, which bond together to form a plastic called polylactide 
(FLA). 

FLA can be used for products such as plant pots and disposable nappies. It has been commercially 
available in some countries since 1990, and certain blends have proved successful in medical implants, sutures 
and drug delivery systems because they are able to dissolve away over time. However, because FLA is much 
more expensive than normal plastics, it has not become as popular as one would have hoped. 

Questions 

1. In your own words, explain what is meant by a 'biodegradable plastic'. 

2. Using your knowledge of chemical bonding, explain why some polymers are biodegradable and others 
are not. 

3. Explain why lactic acid is a more useful monomer than starch, when making a biodegradable plastic. 

4. If you were a consumer (shopper), would you choose to buy a biodegradable plastic rather than another? 
Explain your answer. 

5. What do you think could be done to make biodegradable plastics more popular with consumers? 

2.1.6.3 Plastics and the environment 

Although plastics have had a huge impact globally, there is also an environmental price that has to be paid 
for their use. The following are just some of the ways in which plastics can cause damage to the environment. 

1. Waste disposal Flastics are not easily broken down by micro-organisms and therefore most are not 
easily biodegradeable. This leads to waste dispoal problems. 



42 CHAPTER 2. ORGANIC MACROMOLECULES 

2. Air pollution When plastics burn, they can produce toxic gases such as carbon monoxide, hydrogen 
cyanide and hydrogen chloride (particularly from PVC and other plastics that contain chlorine and 
nitrogen). 

3. Recycling It is very difficult to recycle plastics because each type of plastic has different properties 
and so different recycling methods may be needed for each plastic. However, attempts are being made 
to find ways of recycling plastics more effectively. Some plastics can be remelted and re-used, while 
others can be ground up and used as a filler. However, one of the problems with recycling plastics is 
that they have to be sorted according to plastic type. This process is difficult to do with machinery, and 
therefore needs a lot of labour. Alternatively, plastics should be re-used. In many countries, including 
South Africa, shoppers must now pay for plastic bags. This encourages people to collect and re-use 
the bags they already have. 



2.1.6.3.1 Case Study : Plastic pollution in South Africa 

Read the following extract, taken from 'Planet Ark' (September 2003), and then answer the questions that 
follow. 

South Africa launches a programme this week to exterminate its "national flower " - the millions 
of used plastic bags that litter the landscape. 

Beginning on Friday, plastic shopping bags used in the country must be both thicker and more 
recyclable, a move ofHcials hope will stop people from simply tossing them away. "Government has 
targeted plastic bags because they are the most visible kind of waste, " said Phindile Makwakwa, 
spokeswoman for the Department of Environmental Affairs and Tourism. "But this is mostly 
about changing people's mindsets about the environment." 

South Africa is awash in plastic pollution. Plastic bags are such a common eyesore that they 
are dubbed "roadside daisies" and referred to as the national flower. Bill Naude of the Plastics 
Federation of South Africa said the country used about eight billion plastic bags annually, a figure 
which could drop by 50 percent if the new law works. 

It is difficult sometimes to imagine exactly how much waste is produced in our country every year. Where 
does all of this go to? You are going to do some simple calculations to try to estimate the volume of plastic 
packets that is produced in South Africa every year. 

1. Take a plastic shopping packet and squash it into a tight ball. 

a. Measure the approximate length, breadth and depth of your squashed plastic bag. 

b. Calculate the approximate volume that is occupied by the packet. 

c. Now calculate the approximate volume of your classroom by measuring its length, breadth and 
height. 

d. Calculate the number of squashed plastic packets that would fit into a classroom of this volume. 

e. If South Africa produces an average of 8 billion plastic bags each year, how many clasrooms would 
be filled if all of these bags were thrown away and not re-used? 

2. What has South Africa done to try to reduce the number of plastic bags that are produced? 

3. Do you think this has helped the situation? 

4. What can you do to reduce the amount of plastic that you throw away? 



43 

2.2 Biological macromolecules' 
2.2.1 Biological Macromolecules 

A biological macromolecule is one that is found in living organisms. Biological macromolecules include 
molecules such as carbohydrates, proteins and nucleic acids. Lipids are also biological macromolecules. 
They are essential for all known forms of life to survive. 

Definition 2.4: Biological macromolecule 

A biological macromolecule is a polymer that occurs naturally in living organisms. These molecules 
are essential to the survival of life. 

2.2.1.1 Carbohydrates 

Carbohydrates include the sugars and their polymers. One key characteristic of the carbohydrates is 
that they contain only the elements carbon, hydrogen and oxygen. In the carbohydrate monomers, every 
carbon except one has a hydroxyl group attached to it, and the remaining carbon atom is double bonded to 
an oxygen atom to form a carbonyl group. One of the most important monomers in the carbohydrates is 
glucose (Figure 2.13). The glucose molecule can exist in an open-chain (acyclic) and ring (cyclic) form. 



Image notjintshed 

Figure 2.13: The open chain (a) and cyclic (b) structure of a glucose molecule 



Glucose is produced during photosynthesis, which takes place in plants. During photosynthesis, sunlight 
(solar energy), water and carbon dioxide are involved in a chemical reaction that produces glucose and oxygen. 
This glucose is stored in various ways in the plant. 

The photosynthesis reaction is as follows: 

6CO2 + 6H2O + sunlight -^ CeHi206 + 6O2 

Glucose is an important source of energy for both the plant itself, and also for the other animals and 
organisms that may feed on it. Glucose plays a critical role in cellular respiration, which is a chemical 
reaction that occurs in the cells of all living organisms. During this reaction, glucose and oxygen react to 
produce carbon dioxide, water and Adenosine Triphosphate (ATP). ATP is a molecule that cells use for 
energy so that the body's cells can function normally. The purpose of eating then, is to obtain glucose which 
the body can then convert into the ATP it needs to be able to survive. 

The reaction for cellular respiration is as follows: 

6C6Hi20e + 6O2 ^ 6CO2 + 6H2O + ATP 

We don't often eat glucose in its simple form. More often, we eat complex carbohydrates that our 
bodies have to break down into individual glucose molecules before they can be used in cellular respiration. 
These complex carbohydrates are polymers, which form through condensation polymerisation reactions (Fig- 
ure 2.14). Starch and cellulose are two example of carbohydrates that are polymers composed of glucose 
monomers. 



^This content is available online at <http://siyavula.cnx.Org/content/m39433/l.l/>. 



44 CHAPTER 2. ORGANIC MACROMOLECULES 



Image notjinished 



Figure 2.14: Two glucose monomers (a) undergo a condensation reaction to produce a section of a 
carbohydrate polymer (b). One molecule of water is produced for every two monomers that react. 



Starch Starch is used by plants to store excess glucose, and consists of long chains of glucose monomers. 
Potatoes are made up almost entirely of starch. This is why potatoes are such a good source of energy. 
Animals are also able to store glucose, but in this case it is stored as a compound called glycogen, 
rather than as starch. 

Cellulose Cellulose is also made up of chains of glucose molecules, but the bonding between the 
polymers is slightly different from that in starch. Cellulose is found in the cell walls of plants and is 
used by plants as a building material. 

NOTE: It is very difficult for animals to digest the cellulose in plants that they may have 
been feeding on. However, fungi and some protozoa are able to break down cellulose. Many 
animals, including termites and cows, use these organisms to break cellulose down into glucose, 
which they can then use more easily. 



2.2.1.2 Proteins 

Proteins are an incredibly important part of any cell, and they carry out a number of functions such as 
support, storage and transport within the body. The monomers of proteins are called amino acids. An 
amino acid is an organic molecule that contains a carboxyl and an amino group, as well as a carbon side 
chain. The carbon side chain varies from one amino acid to the next, and is sometimes simply represented 
by the letter 'R' in a molecule's structural formula. Figure 2.15 shows some examples of different amino 
acids. 



Image notjinished 

Figure 2.15: Three amino acids: glycine, alanine and serine 



Although each of these amino acids has the same basic structure, their side chains ('R' groups) are 
different. In the amino acid glycine, the side chain consists only of a hydrogen atom, while alanine has 
a methyl side chain. The 'R' group in serine is CH2 - OH. Amongst other things, the side chains affect 
whether the amino acid is hydrophilic (attracted to water) or hydrophobic (repelled by water). If the side 
chain is polar, then the amino acid is hydrophilic, but if the side chain is non-polar then the amino acid is 
hydrophobic. Glycine and alanine both have non-polar side chains, while serine has a polar side chain. 

2.2.1.2.1 Charged regions in an amino acid 

In an amino acid, the amino group acts as a base because the nitrogen atom has a pair of unpaired electrons 
which it can use to bond to a hydrogen ion. The amino group therefore attracts the hydrogen ion from the 



45 

carboxyl group, and ends up having a charge of +1. The carboxyl group from which the hydrogen ion has 
been taken then has a charge of -1. The amino acid glycine can therefore also be represented as shown in 
the figure below. 



Image notjtnished 

Figure 2.16 



When two amino acid monomers are close together, they may be joined to each other by peptide bonds 
(Figure 2.17) to form a polypeptide chain. . The reaction is a condensation reaction. Polypeptides can 
vary in length from a few amino acids to a thousand or more. The polpeptide chains are then joined to each 
other in different ways to form a protein. It is the sequence of the amino acids in the polymer that gives a 
protein its particular properties. 

The sequence of the amino acids in the chain is known as the protein's primary structure. As the 
chain grows in size, it begins to twist, curl and fold upon itself. The different parts of the polypeptide are 
held together by hydrogen bonds, which form between hydrogen atoms in one part of the chain and oxygen 
or nitrogen atoms in another part of the chain. This is known as the secondary structure of the protein. 
Sometimes, in this coiled helical structure, bonds may form between the side chains (R groups) of the amino 
acids. This results in even more irregular contortions of the protein. This is called the tertiary structure 
of the protein. 



Image notjinished 



Figure 2.17: Two amino acids (glycine and alanine) combine to form part of a polypeptide chain. The 
amino acids are joined by a peptide bond between a carbon atom of one amino acid and a nitrogen atom 
of the other amino acid. 



NOTE: There are twenty different amino acids that exist in nature. All cells, both plant and 
animal, build their proteins from only twenty amino acids. At first, this seems like a very small 
number, especially considering the huge number of different proteins that exist. However, if you 
consider that most proteins are made up of polypeptide chains that contain at least 100 amino 
acids, you will start to realise the endless possible combinations of amino acids that are available. 



2.2.1.2.2 The functions of proteins 

Proteins have a number of functions in living organisms. 

• Structural proteins such as collagen in animal connective tissue and keratin in hair, horns and feather 
quills, all provide support. 

• Storage proteins such as albumin in egg white provide a source of energy. Plants store proteins in their 
seeds to provide energy for the new growing plant. 



46 



CHAPTER 2. ORGANIC MACROMOLECULES 



Transport proteins transport other substances in the body. Haemoglobin in the blood for example, is 

a protein that contains iron. Haemoglobin has an affinity (attraction) for oxygen and so this is how 

oxygen is transported around the body in the blood. 

Hormonal proteins coordinate the body's activities. Insulin for example, is a hormonal protein that 

controls the sugar levels in the blood. 

Enzymes are chemical catalysts and speed up chemical reactions. Digestive enzymes such as amylase 

in your saliva, help to break down polymers in food. Enzymes play an important role in all cellular 

reactions such as respiration, photosynthesis and many others. 



2.2.1.2.2.1 Research Project : Macromolecules in our daily diet 

1. In order to keep our bodies healthy, it is important that we eat a balanced diet with the right amounts 
of carbohydrates, proteins and fats. Fats are an important source of energy, they provide insulation 
for the body, and they also provide a protective layer around many vital organs. Our bodies also 
need certain essential vitamins and minerals. Most food packaging has a label that provides this 
information. Choose a number of different food items that you eat. Look at the food label for each, 
and then complete the following table: 



Food 


Carbohydrates (%) 


Proteins (%) 


Fats (%) 











































Table 2.4 

a. Which food type contains the largest proportion of protein? 

b. Which food type contains the largest proportion of carbohydrates? 

c. Which of the food types you have listed would you consider to be the 'healthiest'? Give a reason 
for your answer. 

2. In an effort to lose weight, many people choose to diet. There are many diets on offer, each of which is 
based on particular theories about how to lose weight most effectively. Look at the list of diets below: 

• Vegetarian diet 

• Low fat diet 

• Atkin's diet 

• Weight Watchers 

For each of these diets, answer the following questions: 

a. What theory of weight loss does each type of diet propose? 

b. What are the benefits of the diet? 

c. What are the potential problems with the diet? 



2.2.1.2.2.2 Carbohydrates and proteins 

1. Give the structural formula for each of the following: 

a. A polymer chain, consisting of three glucose molecules. 



47 

b. A polypeptide chain, consisting of two molecules of alanine and one molecule of serine. 

2. Write balanced equations to show the polymerisation reactions that produce the polymers described 
above. 

3. The following polypeptide is the end product of a polymerisation reaction: 



Image notjinished 

Figure 2.18 



a. Give the structural formula of the monomers that make up the polypeptide. 

b. On the structural formula of the first monomer, label the amino group and the carboxyl group. 

c. What is the chemical formula for the carbon side chain in the second monomer? 

d. Name the bond that forms between the monomers of the polypeptide. 



2.2.1.3 Nucleic Acids 

You will remember that we mentioned earlier that each protein is different because of its unique sequence of 
amino acids. But what controls how the amino acids arrange themselves to form the specific proteins that 
are needed by an organism? This task is for the gene. A gene contains DNA (deoxyribonucleic acid) which 
is a polymer that belongs to a class of compounds called the nucleic acids. DNA is the genetic material 
that organisms inherit from their parents. It is DNA that provides the genetic coding that is needed to form 
the specific proteins that an organism needs. Another nucleic acid is RNA (ribonucleic acid). The diagram 
in Figure 2.19 shows an RNA molecule. 

The DNA polymer is made up of monomers called nucleotides. Each nucleotide has three parts: a 
sugar, a phosphate and a nitrogenous base. DNA is a double-stranded helix (a helix is basically a coil). Or 
you can think of it as two RNA molecules bonded together. 



Image notjinished 

Figure 2.19: Nucleotide monomers make up the RNA polymer 



There are five different nitrogenous bases: adenine (A), guanine (G), cytosine (C), thymine (T) and 
uracil (U). It is the sequence of the nitrogenous bases in a DNA polymer that will determine the genetic 
code for that organism. Three consecutive nitrogenous bases provide the coding for one amino acid. So, for 
example, if the nitrogenous bases on three nucleotides are uracil, cytosine and uracil (in that order), one 
serine amino acid will become part of the polypeptide chain. The polypeptide chain is built up in this way 
until it is long enough (and with the right amino acid sequence) to be a protein. Since proteins control much 
of what happens in living organisms, it is easy to see how important nucleic acids are as the starting point 
of this process. 

NOTE: A single defect in even one nucleotide, can be devastating to an organism. One example 
of this is a disease called sickle cell anaemia. Because of one wrong nucletide in the genetic 



48 



CHAPTER 2. ORGANIC MACROMOLECULES 



code, the body produces a protein called sickle haemoglobin. Haemoglobin is the protein in red 
blood cells that helps to transport oxygen around the body. When sickle haemoglobin is produced, 
the red blood cells change shape. This process damages the red blood cell membrane, and can 
cause the cells to become stuck in blood vessels. This then means that the red blood cells, whcih 
are carrying oxygen, can't get to the tissues where they are needed. This can cause serious organ 
damage. Individuals who have sickle cell anaemia generally have a lower life expectancy. 

Table 2.5 shows some other examples of genetic coding for different amino acids. 



Nitrogenous base sequence 


Amino acid 


UUU 


Phenylalanine 


CUU 


Leucine 


UCU 


Serine 


UAU 


Tyrosine 


UGU 


Cysteine 


GUU 


Valine 


GCU 


Alanine 


GGU 


Glycine 



Table 2.5: Nitrogenouse base sequences and the corresponding amino acid 



2.2.1.3.1 Nucleic acids 

1. For each of the following, say whether the statement is true or false. If the statement is false, give a 
reason for your answer. 

a. Deoxyribonucleic acid (DNA) is an example of a polymer and a nucleotide is an example of a 
monomer. 

b. Thymine and uracil are examples of nucleotides. 

c. A person's DNA will determine what proteins their body will produce, and therefore what char- 
acteristics they will have. 

d. An amino acid is a protein monomer. 

e. A polypeptide that consists of five amino acids, will also contain five nucleotides. 

2. For each of the following sequences of nitrogenous bases, write down the amino acid/s that will be part 
of the polypeptide chain. 

a. UUU 

b. UCUUUU 

c. GGUUAUGUUGGU 

3. A polypeptide chain consists of three amino acids. The sequence of nitrogenous bases in the nucleotides 
of the DNA is GCUGGUGCU. Give the structural formula of the polypeptide. 



2.2.2 Summary 



• A polymer is a macromolecule that is made up of many repeating structural units called monomers 
which are joined by covalent bonds. 

• Polymers that contain carbon atoms in the main chain are called organic polymers. 



49 

Organic polymers can be divided into natural organic polymers (e.g. natural rubber) or synthetic 

organic polymers (e.g. polystyrene). 

The polymer polyethene for example, is made up of many ethene monomers that have been joined 

into a polymer chain. 

Polymers form through a process called polymerisation. 

Two examples of polymerisation reactions are addition and condensation reactions. 

An addition reaction occurs when unsaturated monomers (e.g. alkenes) are added to each other one 

by one. The breaking of a double bond between carbon atoms in the monomer, means that a bond 

can form with the next monomer. The polymer polyethene is formed through an addition reaction. 

In a condensation reaction, a molecule of water is released as a product of the reaction. The water 

molecule is made up of atoms that have been lost from each of the monomers. Polyesters and nylon 

are polymers that are produced through a condensation reaction. 

The chemical properties of polymers (e.g. tensile strength and melting point) are determined by 

the types of atoms in the polymer, and by the strength of the bonds between adjacent polymer chains. 

The stronger the bonds, the greater the strength of the polymer, and the higher its melting point. 

One group of synthetic organic polymers, are the plastics. 

Polystyrene is a plastic that is made up of styrene monomers. Polystyrene is used a lot in packaging. 

Polyvinyl chloride (PVC) consists of vinyl chloride monomers. PVC is used to make pipes and 

flooring. 

Polyethene, or polyethylene, is made from ethene monomers. Polyethene is used to make film 

wrapping, plastic bags, electrical insulation and bottles. 

Polytetrafluoroethylene is used in non-stick frying pans and electrical insulation. 

A thermoplastic can be heated and melted to a liquid. It freezes to a brittle, glassy state when cooled 

very quickly. Examples of thermoplastics are polyethene and PVC. 

A thermoset plastic cannot be melted or re-shaped once formed. Examples of thermoset plastics are 

vulcanised rubber and melanine. 

It is not easy to recycle all plastics, and so they create environmental problems. 

Some of these environmental problems include issues of waste disposal, air pollution and recycling. 

A biological macromolecule is a polymer that occurs naturally in living organisms. 
Examples of biological macromolecules include carbohydrates and proteins, both of which are es- 
sential for life to survive. 

Carbohydrates include the sugars and their polymers, and are an important source of energy in living 
organisms. 
Glucose is a carbohydrate monomer. Glucose is the molecule that is needed for cellular respiration. 

The glucose monomer is also a building block for carbohydrate polymers such as starch, glycogen 

and cellulose. 

Proteins have a number of important functions. These include their roles in structures, transport, 

storage, hormonal proteins and enzymes. 

A protein consists of monomers called amino acids, which are joined by peptide bonds. 

A protein has a primary, secondary and tertiary structure. 

An amino acid is an organic molecule, made up of a carboxyl and an amino group, as well as a 

carbon side chain of varying lengths. 

It is the sequence of amino acids that determines the nature of the protein. 

It is the DNA of an organism that determines the order in which amino acids combine to make a 

protein. 

DNA is a nucleic acid. DNA is a polymer, and is made up of monomers called nucleotides. 

Each nucleotide consists of a sugar, a phosphate and a nitrogenous base. It is the sequence of the 

nitrogenous bases that provides the 'code' for the arrangement of the amino acids in a protein. 



50 CHAPTER 2. ORGANIC MACROMOLECULES 

2.2.2.1 Summary exercise 

1. Give one word for each of the following descriptions: 

a. A chain of monomers joined by covalent bonds. 

b. A polymerisation reaction that produces a molecule of water for every two monomers that bond. 

c. The bond that forms between an alcohol and a carboxylic acid monomer during a polymerisation 
reaction. 

d. The name given to a protein monomer. 

e. A six-carbon sugar monomer. 

f. The monomer of DNA, which determines the sequence of amino acids that will make up a protein. 

2. For each of the following questions, choose the one correct answer from the list provided. 

a. A polymer is made up of monomers, each of which has the formula CH2=CHCN. The formula of 
the polymer is: 

1. -(CH2=CHCN)„- 

2. -(CH2-CHCN)„- 

3. -(CH-CHCN)„- 

4. -(CH3-CHCN)„- 

b. A polymer has the formula -[CO(CH2)4CO-NH(CH2)6NH]„-. Which of the following statements 
is true? 

1. The polymer is the product of an addition reaction. 

2. The polymer is a polyester. 

3. The polymer contains an amide linkage. 

4. The polymer contains an ester linkage. 

c. Glucose... 

1. is a monomer that is produced during cellular respiration 

2. is a sugar polymer 

3. is the monomer of starch 

4. is a polymer produced during photosynthesis 

3. The following monomers are involved in a polymerisation reaction: 



Image notjinished 

Figure 2.20 



a. Give the structural formula of the polymer that is produced. 

b. Is the reaction an addition or condensation reaction? 

c. To what group of organic compounds do the two monomers belong? 

d. What is the name of the monomers? 

e. What type of bond forms between the monomers in the final polymer? 

4. The table below shows the melting point for three plastics. Suggest a reason why the melting point of 
PVC is higher than the melting point for polyethene, but lower than that for polyester. 



51 



Plastic 


Melting point (°C) 


Polyethene 


105 - 115 


PVC 


212 


Polyester 


260 



Table 2.6 



5. An amino acid has the formula H2NCH(CH2CH2SCH3)COOH. 

a. Give the structural formula of this amino acid. 

b. What is the chemical formula of the carbon side chain in this molecule? 

c. Are there any peptide bonds in this molecule? Give a reason for your answer. 



52 CHAPTER 2. ORGANIC MACROMOLECULES 

Solutions to Exercises in Chapter 2 

Solution to Exercise 2.1 (p. 34) 

Step 1. The monomer is: 

Image notjinished 

Figure 2.21 



Step 2. The monomer has a double bond between two carbon atoms. The monomer must be an alkene. 
Step 3. In this example, unsaturated monomers combine to form a saturated polymer. No atoms are lost or 

gained for the bonds between monomers to form. They are simply added to each other. This is an 

addition reaction. 



Chapter 3 

Reaction rates 

3.1 Introduction' 

3.1.1 Introduction 

Before we begin this section, it might be useful to think about some different types of reactions and how 
quickly or slowly they occur. 

3.1.1.1 Thinking about reaction rates 

Think about each of the following reactions: 



rusting of metals 

photosynthesis 

weathering of rocks (e.g. limestone rocks being weathered by water) 



• combustion 

1. For each of the reactions above, write a chemical equation for the reaction that takes place. 

2. How fast is each of these reactions? Rank them in order from the fastest to the slowest. 

3. How did you decide which reaction was the fastest and which was the slowest? 

4. Try to think of some other examples of chemical reactions. How fast or slow is each of these reactions, 
compared with those listed earlier? 

In a chemical reaction, the substances that are undergoing the reaction are called the reactants, while the 
substances that form as a result of the reaction are called the products. The reaction rate describes how 
quickly or slowly the reaction takes place. So how do we know whether a reaction is slow or fast? One way 
of knowing is to look either at how quickly the reactants are used up during the reaction or at how quickly 
the product forms. For example, iron and sulfur react according to the following equation: 

Fe + S ^ FeS 

In this reaction, we can observe the speed of the reaction by measuring how long it takes before there is 
no iron or sulfur left in the reaction vessel. In other words, the reactants have been used up. Alternatively, 
one could see how quickly the iron sulfide product forms. Since iron sulfide looks very different from either 
of its reactants, this is easy to do. 

In another example: 

2Mg{s) + 02^2MgO{s) (3.1) 



^This content is available online at <http://siyavula.cnx.Org/content/m39459/l.l/>. 

53 



54 



CHAPTER 3. REACTION RATES 



In this case, the reaction rate depends on the speed at which the reactants (soHd magnesium and oxygen 
gas) are used up, or the speed at which the product (magnesium oxide) is formed. 

Definition 3.1: Reaction rate 

The rate of a reaction describes how quickly reactants are used up or how quickly products are 
formed during a chemical reaction. The units used are: moles per second (mols/second or mol.s"^). 

The average rate of a reaction is expressed as the number of moles of reactant used up, divided by the 
total reaction time, or as the number of moles of product formed, divided by the reaction time. Using the 
magnesium reaction shown earlier: 



Average reaction rate of Mg 



moles Mg used 
reaction time (s) 



(3.2) 



Average reaction rate of O2 



moles O2 used 
reaction time (s) 



(3.3) 



or 



Average reaction rate of MgO 



moles MgO produced 
reaction time (s) 



(3.4) 



Exercise 3.1: Reaction rates (Solution on p. 78.) 

The following reaction takes place: 
ALi + 02^ 2Li20 
After two minutes , 4 g of Lithium has been used up. Calculate the rate of the reaction. 



3.1.1.2 Reaction rates 

1. A number of different reactions take place. The table below shows the number of moles of reactant 
that are used up in a particular time for each reaction. 



Reaction 


Mols used up 


Time 


Reaction rate 


1 


2 


30 sees 




2 


5 


2 mins 




3 


1 


1.5 mins 




4 


3.2 


1.5 mins 




5 


5.9 


30 sees 





Table 3.1 



a. Complete the table by calculating the rate of each reaction. 

b. Which is the fastest reaction? 

c. Which is the slowest reaction? 

2. Two reactions occur simultaneously in separate reaction vessels. The reactions are as follows: Mg + 
CI2 -^ MgCl22Na + CI2 -^ 2NaCl After 1 minute, 2 g of MgCl2 have been produced in the first 
reaction. 

a. How many moles of MgCl2 are produced after 1 minute? 

b. Calculate the rate of the reaction, using the amount of product that is produced. 



55 



c. Assuming that the second reaction also proceeds at the same rate, calculate... 

1. the number of moles of NaCl produced after 1 minute. 

2. the mass (in g) of sodium that is needed for this reaction to take place. 



3.1.2 Factors afFecting reaction rates 

Several factors affect the rate of a reaction. It is important to know these factors so that reaction rates can 
be controlled. This is particularly important when it comes to industrial reactions, so that productivity can 
be maximised. The following are some of the factors that affect the rate of a reaction. 

1. Nature of reactants Substances have different chemical properties and therefore react differently 
and at different rates. 

2. Concentration (or pressure in the case of gases) As the concentration of the reactants increases, so 
does the reaction rate. 

3. Temperature If the temperature of the reaction increases, so does the rate of the reaction. 

4. Catalyst Adding a catalyst increases the reaction rate. 

5. Surface area of solid reactants Increasing the surface area of the reactants (e.g. if a solid reactant 
is broken or ground up into smaller pieces) will increase the reaction rate. 



3.1.2.1 Experiment : The nature of reactants. 

Aim: 

To determine the effect of the nature of reactants on the rate of a reaction. 
Apparatus: 

Oxalic acid ((C00H)2), iron(II) sulphate (FeS04), potassium permanganate (KMn04), concentrated 
sulfuric acid (H2SO4), spatula, test tubes, medicine dropper, glass beaker and glass rod. 



Image notjinished 

Figure 3.1 



Method: 

1. In the first test tube, prepare an iron (II) sulphate solution by dissolving about two spatula points of 
iron (II) sulphate in 10 cm^ of water. 

2. In the second test tube, prepare a solution of oxalic acid in the same way. 

3. Prepare a solution of sulfuric acid by adding 1 cm"^ of the concentrated acid to about 4 cm'^ of water. 
Remember always to add the acid to the water, and never the other way around. 

4. Add 2 cm^ of the sulfuric acid solution to the iron(II) and oxalic acid solutions respectively. 

5. Using the medicine dropper, add a few drops of potassium permanganate to the two test tubes. Once 
you have done this, observe how quickly each solution discolours the potassium permanganate solution. 

Results: 

• You should have seen that the oxalic acid solution discolours the potassium permanganate much more 
slowly than the iron(II) sulphate. 



56 CHAPTER 3. REACTION RATES 

• It is the oxalate ions (€204^) and the Fe^"*" ions that cause the discolouration. It is clear that the Fe^+ 
ions act much more quickly than the €204" ions. The reason for this is that there are no covalent 
bonds to be broken in the ions before the reaction can take place. In the case of the oxalate ions, 
covalent bonds between carbon and oxygen atoms must be broken first. 

Conclusions: 

The nature of the reactants can affect the rate of a reaction. 

NOTE: Oxalic acids are abundant in many plants. The leaves of the tea plant {Camellia sinensis) 
contain very high concentrations of oxalic acid relative to other plants. Oxalic acid also occurs in 
small amounts in foods such as parsley, chocolate, nuts and berries. Oxalic acid irritates the lining 
of the gut when it is eaten, and can be fatal in very large doses. 



3.1.2.2 Experiment : Surface area and reaction rates. 

Marble (CaCOs) reacts with hydrochloric acid (HCl) to form calcium chloride, water and carbon dioxide 
gas according to the following equation: 

CaCOa + 2HCI -^ CaCh + H2O + CO2 

Aim: 

To determine the effect of the surface area of reactants on the rate of the reaction. 

Apparatus: 

2 g marble chips, 2 g powdered marble, hydrochloric acid, beaker, two test tubes. 



Image notjinished 

Figure 3.2 



Method: 

1. Prepare a solution of hydrochloric acid in the beaker by adding 2 cm^ of the concentrated solution to 
20 cm^ of water. 

2. Place the marble chips and powdered marble into separate test tubes. 

3. Add 10 cm'^ of the dilute hydrochloric acid to each of the test tubes and observe the rate at which 
carbon dioxide gas is produced. 

Results: 

• Which reaction proceeds the fastest? 

• Can you explain this? 

Conclusions: 

The reaction with powdered marble is the fastest. The smaller the pieces of marble are, the greater the 
surface area for the reaction to take place. The greater the surface area of the reactants, the faster the 
reaction rate will be. 



57 

3.1.2.3 Experiment : Reactant concentration and reaction rate. 

Aim: 

To determine the effect of reactant concentration on reaction rate. 
Apparatus: 

Concentrated hydrochloric acid (HCl), magnesium ribbon, two beakers, two test tubes, measuring cyHn- 
der. 

Method: 

1. Prepare a solution of dilute hydrochloric acid in one of the beakers by diluting 1 part concentrated acid 
with 10 parts water. For example, if you measure 1 cm"^ of concentrated acid in a measuring cylinder 
and pour it into a beaker, you will need to add 10 cm"^ of water to the beaker as well. In the same way, 
if you pour 2 cm^ of concentrated acid into a beaker, you will need to add 20 cm'^ of water. Both of 
these are 1:10 solutions. Pour 10 cm"^ of the 1:10 solution into a test tube and mark it 'A'. Remember 
to add the acid to the water, and not the other way around. 

2. Prepare a second solution of dilute hydrochloric acid by diluting 1 part concentrated acid with 20 parts 
water. Pour lOcm^ of this 1:20 solution into a second test tube and mark it 'B'. 

3. Take two pieces of magnesium ribbon of the same length. At the same time, put one piece of 
magnesium ribbon into test tube A and the other into test tube B, and observe closely what happens. 



Image notjinished 

Figure 3.3 



The equation for the reaction is: 
2HCI + Mg^ MgCh + ^2 
Results: 

• Which of the two solutions is more concentrated, the 1:10 or 1:20 hydrochloric acid solution? 

• In which of the test tubes is the reaction the fastest? Suggest a reason for this. 

• How can you measure the rate of this reaction? 

• What is the gas that is given off? 

• Why was it important that the same length of magnesium ribbon was used for each reaction? 

Conclusions: 

The 1:10 solution is more concentrated and this reaction therefore proceeds faster. The greater the 
concentration of the reactants, the faster the rate of the reaction. The rate of the reaction can be measured 
by the rate at which hydrogen gas is produced. 

3.1.2.4 Group work : The effect of temperature on reaction rate 

1. In groups of 4-6, design an experiment that will help you to see the effect of temperature on the reaction 
time of 2 cm of magnesium ribbon and 20 ml of vinegar. During your group discussion, you should 
think about the following: 

• What equipment will you need? 

• How will you conduct the experiment to make sure that you are able to compare the results for 
different temperatures? 

• How will you record your results? 



58 CHAPTER 3. REACTION RATES 

• What safety precautions will you need to take when you carry out this experiment? 

2. Present your experiment ideas to the rest of the class, and give them a chance to comment on what 
you have done. 

3. Once you have received feedback, carry out the experiment and record your results. 

4. What can you conclude from your experiment? 



3.2 Collision theory, measurement and mechanism' 

3.2,1 Reaction rates and collision theory 

It should be clear now that the rate of a reaction varies depending on a number of factors. But how can we 
explain why reactions take place at different speeds under different conditions? Why, for example, does an 
increase in the surface area of the reactants also increase the rate of the reaction? One way to explain this 
is to use collision theory. 

For a reaction to occur, the particles that are reacting must collide with one another. Only a fraction 
of all the collisions that take place actually cause a chemical change. These are called 'successful' collisions. 
When there is an increase in the concentration of reactants, the chance that reactant particles will collide 
with each other also increases because there are more particles in that space. In other words, the collision 
frequency of the reactants increases. The number of successful collisions will therefore also increase, and so 
will the rate of the reaction. In the same way, if the surface area of the reactants increases, there is also a 
greater chance that successful collisions will occur. 

Definition 3.2: Collision theory 

Collision theory is a theory that explains how chemical reactions occur and why reaction rates 
differ for different reactions. The theory states that for a reaction to occur the reactant particles 
must collide, but that only a certain fraction of the total collisions, the effective collisions, actually 
cause the reactant molecules to change into products. This is because only a small number of the 
molecules have enough energy and the right orientation at the moment of impact to break the 
existing bonds and form new bonds. 

When the temperature of the reaction increases, the average kinetic energy of the reactant particles 
increases and they will move around much more actively. They are therefore more likely to collide with 
one another (Figure 3.4). Increasing the temperature also increases the number of particles whose energy 
will be greater than the activation energy for the reaction (refer "Mechanism of reaction and catalysis" 
(Section 3.2.3: Mechanism of reaction and catalysis)). 



Image notjinished 



Figure 3.4: An increase in the temperature of a reaction increases the chances that the reactant particles 
(A and B) will collide because the particles have more energy and move around more. 



^This content is available online at <http://siyavula.cnx.Org/content/m39465/l.l/>. 



59 



Image notjinished 



Figure 3.5 



run demo^ 



3.2.1.1 Rates of reaction 



Hydrochloric acid and calcium carbonate react according to the following equation: 

CaCOs + 2HCI -^ CaCh + H^O + CO2 

The volume of carbon dioxide that is produced during the reaction is measured at different times. The 
results are shown in the table below. 



Time (mins) 


Total Volume of CO2 produced (cm'^) 


1 


14 


2 


26 


3 


36 


4 


44 


5 


50 


6 


58 


7 


65 


8 


70 


9 


74 


10 


77 



Table 3.2 

Note: On a graph of production against time, it is the gradient of the graph that shows the rate of the 
reaction. 

Questions: 

1. Use the data in the table to draw a graph showing the volume of gas that is produced in the reaction, 
over a period of 10 minutes. 

2. At which of the following times is the reaction fastest? Time = 1 minute; time = 6 minutes or time = 
8 minutes. 

3. Suggest a reason why the reaction slows down over time. 

4. Use the graph to estimate the volume of gas that will have been produced after 11 minutes. 

5. After what time do you think the reaction will stop? 

6. If the experiment was repeated using a more concentrated hydrochloric acid solution... 

a. would the rate of the reaction increase or decrease from the one shown in the graph? 

b. draw a rough line on the graph to show how you would expect the reaction to proceed with a 
more concentrated HCl solution. 



^ http://phet.colorado.edu/sims/reactions-and-rates/re actions- and-ratesen.jnlp 



60 CHAPTER 3. REACTION RATES 

3.2,2 Measuring Rates of Reaction 

How the rate of a reaction is measured will depend on what the reaction is, and what product forms. Look 
back to the reactions that have been discussed so far. In each case, how was the rate of the reaction measured? 
The following examples will give you some ideas about other ways to measure the rate of a reaction: 

• Reactions that produce hydrogen gas: When a metal dissolves in an acid, hydrogen gas is produced. A 
lit splint can be used to test for hydrogen. The 'pop' sound shows that hydrogen is present. For example, 
magnesium reacts with sulfuric acid to produce magnesium sulphate and hydrogen. Mg {s) + H2S04 -^ 
MgSOi + H2 

• Reactions that produce carbon dioxide: When a carbonate dissolves in an acid, carbon dioxide gas is 
produced. When carbon dioxide is passes through limewater, it turns the limewater milky. This is a 
simple test for the presence of carbon dioxide. For example, calcium carbonate reacts with hydrochloric 
acid to produce calcium chloride, water and carbon dioxide. C0CO3 (s) + 2HCI {aq) -^ CaCl2 {aq) + 

H2O (0 + CO2 (5) 

• Reactions that produce gases such as oxygen or carbon dioxide: Hydrogen peroxide decomposes to 
produce oxygen. The volume of oxygen produced can be measured using the gas syringe method 
(Figure 3.6). The gas collects in the syringe, pushing out against the plunger. The volume of gas 
that has been produced can be read from the markings on the syringe. For example, hydrogen per- 
oxide decomposes in the presence of a manganese (IV) oxide catalyst to produce oxygen and water. 
2H2O2 {aq) -^ 2H2O (0 + O2 (5) 



Image notjtnished 

Figure 3.6: Gas Syringe Method 



Precipitate reactions: In reactions where a precipitate is formed, the amount of precipitate formed in a 
period of time can be used as a measure of the reaction rate. For example, when sodium thiosulphate 
reacts with an acid, a yellow precipitate of sulfur is formed. The reaction is as follows: Na2S20z {aq) + 
2HCI {aq) -^ 2NaCl {aq) + 5*02 {aq) + H2O {I) + S (s) One way to estimate the rate of this reaction 
is to carry out the investigation in a conical flask and to place a piece of paper with a black cross 
underneath the bottom of the flask. At the beginning of the reaction, the cross will be clearly visible 
when you look into the flask (Figure 3.7). However, as the reaction progresses and more precipitate is 
formed, the cross will gradually become less clear and will eventually disappear altogether. Noting the 
time that it takes for this to happen will give an idea of the reaction rate. Note that it is not possible 
to collect the SO2 gas that is produced in the reaction, because it is very soluble in water. 



Image notjtnished 



Figure 3.7: At the beginning of the reaction beteen sodium thiosulphate and hydrochloric acid, when 
no precipitate has been formed, the cross at the bottom of the conical flask can be clearly seen. 



Changes in mass: The rate of a reaction that produces a gas can also be measured by calculating 
the mass loss as the gas is formed and escapes from the reaction flask. This method can be used for 
reactions that produce carbon dioxide or oxygen, but are not very accurate for reactions that give off 
hydrogen because the mass is too low for accuracy. Measuring changes in mass may also be suitable 
for other types of reactions. 



61 

3.2.2.1 Experiment : Measuring reaction rates 

Aim: 

To measure the effect of concentration on the rate of a reaction. 
Apparatus: 

• 300 cm^ of sodium thiosulphate (Na2S203) solution. Prepare a solution of sodium thiosulphate by 
adding 12 g of Na2S203 to 300 cm^ of water. This is solution 'A'. 

• 300 cm"^ of water 

• 100 cm'^ of 1:10 dilute hydrochloric acid. This is solution 'B'. 

• Six 100 cm"^ glass beakers 

• Measuring cylinders 

• Paper and marking pen 

• Stopwatch or timer 

Method: 

One way to measure the rate of this reaction is to place a piece of paper with a cross underneath the 
reaction beaker to see how quickly the cross is made invisible by the formation of the sulfur precipitate. 

1. Set up six beakers on a fiat surface and mark them from 1 to 6. Under each beaker you will need to 
place a piece of paper with a large black cross. 

2. Pour 60 cm"^ solution A into the first beaker and add 20 cm^ of water 

3. Use the measuring cylinder to measure 10 cm^ HCl. Now add this HCl to the solution that is already 
in the first beaker (NB: Make sure that you always clean out the measuring cylinder you have used 
before using it for another chemical) . 

4. Using a stopwatch with seconds, record the time it takes for the precipitate that forms to block out 
the cross. 

5. Now measure 50 cm^ of solution A into the second beaker and add 30 cm"^ of water. To this second 
beaker, add 10 cm'^ HCl, time the reaction and record the results as you did before. 

6. Continue the experiment by diluting solution A as shown below. 



Beaker 


Solution A (cm^) 


Water (cm'^) 


Solution B (cm^) 


Time (s) 


1 


60 


20 


10 




2 


50 


30 


10 




3 


40 


40 


10 




4 


30 


50 


10 




5 


20 


60 


10 




6 


10 


70 


10 





Table 3.3 

The equation for the reaction between sodium thiosulphate and hydrochloric acid is: 

Na2S203 {aq) + 2HCI {aq) -^ 2NaCl [aq) + SO2 [aq) + H2O [1) + S (s) 

Results: 

• Calculate the reaction rate in each beaker. This can be done using the following equation: 

1 



Rate of reaction 



time 



(3.5) 



62 CHAPTER 3. REACTION RATES 

• Represent your results on a graph. Concentration will be on the x-axis and reaction rate on the 

y-axis. Note that the original volume of Na2S203 can be used as a measure of concentration. 

• Why was it important to keep the volume of HCl constant? 

• Describe the relationship between concentration and reaction rate. 

Conclusions: 

The rate of the reaction is fastest when the concentration of the reactants was the highest. 

3.2,3 Mechanism of reaction and catalysis 

Earlier it was mentioned that it is the collision of particles that causes reactions to occur and that only 
some of these collisions are 'successful'. This is because the reactant particles have a wide range of kinetic 
energy, and only a small fraction of the particles will have enough energy to actually break bonds so that a 
chemical reaction can take place. The minimum energy that is needed for a reaction to take place is called 
the activation energy. For more information on the energy of reactions, refer to Grade 11. 

Definition 3.3: Activation energy 

The energy that is needed to break the bonds in reactant molecules so that a chemical reaction 
can proceed. 

Even at a fixed temperature, the energy of the particles varies, meaning that only some of them will have 
enough energy to be part of the chemical reaction, depending on the activation energy for that reaction. 
This is shown in Figure 3.8. Increasing the reaction temperature has the effect of increasing the number of 
particles with enough energy to take part in the reaction, and so the reaction rate increases. 



Image notjtnished 

Figure 3.8: The distribution of particle kinetic energies at a fixed temperature 



A catalyst functions slightly differently. The function of a catalyst is to lower the activation energy 
so that more particles now have enough energy to react. The catalyst itself is not changed during the 
reaction, but simply provides an alternative pathway for the reaction, so that it needs less energy. Some 
metals e.g. platinum, copper and iron can act as catalysts in certain reactions. In our own human bodies, 
enzymes are catalysts that help to speed up biological reactions. Catalysts generally react with one or more 
of the reactants to form a chemical intermediate which then reacts to form the final product. The chemical 
intermediate is sometimes called the activated complex. 

The following is an example of how a reaction that involves a catalyst might proceed. C represents the 
catalyst, A and B are reactants and D is the product of the reaction of A and B. 

Step 1: A + C ^ AC 

Step 2: B + AC ^ ABC 

Step 3: ABC -^ CD 

Step 4: CD ^ C + D 

In the above, ABC represents the intermediate chemical. Although the catalyst (C) is consumed by 
reaction 1, it is later produced again by reaction 4, so that the overall reaction is as follows: 

A+B+C^D+C 

You can see from this that the catalyst is released at the end of the reaction, completely unchanged. 



63 

Definition 3.4: Catalyst 

A catalyst speeds up a chemical reaction, without being altered in any way. It increases the 
reaction rate by lowering the activation energy for a reaction. 

Energy diagrams are useful to illustrate the effect of a catalyst on reaction rates. Catalysts decrease the 
activation energy required for a reaction to proceed (shown by the smaller 'hump' on the energy diagram in 
Figure 3.9), and therefore increase the reaction rate. 

Image notjinished 

Figure 3.9: The effect of a catalyst on the activation energy of a reaction 

3.2.3.1 Experiment : Catalysts and reaction rates 

Aim: 

To determine the effect of a catalyst on the rate of a reaction 

Apparatus: 

Zinc granules, 0.1 M hydrochloric acid, copper pieces, one test tube and a glass beaker. 

Method: 

1. Place a few of the zinc granules in the test tube. 

2. Measure the mass of a few pieces of copper and keep them separate from the rest of the copper. 

3. Add about 20 cm^ of HCl to the test tube. You will see that a gas is released. Take note of how quickly 
or slowly this gas is released. Write a balanced equation for the chemical reaction that takes place. 

4. Now add the copper pieces to the same test tube. What happens to the rate at which the gas is 
produced? 

5. Carefully remove the copper pieces from the test tube (do not get HCl on your hands), rinse them in 
water and alcohol and then weigh them again. Has the mass of the copper changed since the start of 
the experiment? 

Results: 

During the reaction, the gas that is released is hydrogen. The rate at which the hydrogen is produced 
increases when the copper pieces (the catalyst) are added. The mass of the copper does not change during 
the reaction. 

Conclusions: 

The copper acts as a catalyst during the reaction. It speeds up the rate of the reaction, but is not changed 
in any way itself. 

3.2.3.2 Reaction rates 

1. For each of the following, say whether the statement is true or false. If it is false, re- write the 
statement correctly. 

a. A catalyst increases the energy of reactant molecules so that a chemical reaction can take place. 

b. Increasing the temperature of a reaction has the effect of increasing the number of reactant 
particles that have more energy that the activation energy. 

c. A catalyst does not become part of the final product in a chemical reaction. 



64 CHAPTER 3. REACTION RATES 

2. 5 g of zinc granules are added to 400 cm^ of 0.5 mol.dm"^ hydrochloric acid. To investigate the rate 
of the reaction, the change in the mass of the flask containing the zinc and the acid was measured 
by placing the flask on a direct reading balance. The reading on the balance shows that there is a 
decrease in mass during the reaction. The reaction which takes place is given by the following equation: 
Zn (s) + 2HCI {aq) -^ ZnCh {aq) + H2 (ff) 

a. Why is there a decrease in mass during the reaction? 

b. The experiment is repeated, this time using 5 g of powdered zinc instead of granulated zinc. How 
will this influence the rate of the reaction? 

c. The experiment is repeated once more, this time using 5 g of granulated zinc and 600 cm^ of 0.5 
mol.dm""^ hydrochloric acid. How does the rate of this reaction compare to the original reaction 
rate? 

d. What effect would a catalyst have on the rate of this reaction? 

(lEB Paper 2 2003) 

3. Enzymes are catalysts. Conduct your own research to find the names of common enzymes in the human 
body and which chemical reactions they play a role in. 

4. 5 g of calcium carbonate powder reacts with 20 cm^ of a 0.1 mol.dm""^ solution of hydrochloric acid. 
The gas that is produced at a temperature of 25"'''^C is collected in a gas syringe. 

a. Write a balanced chemical equation for this reaction. 

b. The rate of the reaction is determined by measuring the volume of gas thas is produced in the 
first minute of the reaction. How would the rate of the reaction be affected if: 

1. a lump of calcium carbonate of the same mass is used 

2. 40 cm'^ of 0.1 mol.dm"^ hydrochloric acid is used 



3.3 Chemical equilibrium* 

3.3.1 Chemical equilibrium 

Having looked at factors that affect the rate of a reaction, we now need to ask some important questions. 
Does a reaction always proceed in the same direction or can it be reversible? In other words, is it always 
true that a reaction proceeds from reactants to products, or is it possible that sometimes, the reaction will 
reverse and the products will be changed back into the reactants? And does a reaction always run its full 
course so that all the reactants are used up, or can a reaction reach a point where reactants are still present, 
but there does not seem to be any further change taking place in the reaction? The following demonstration 
might help to explain this. 

3.3.1.1 Demonstration : Liquid-vapour phase equilibrium 

Apparatus and materials: 

2 beakers; water; bell jar 
Method: 

1. Half fill two beakers with water and mark the level of the water in each case. 

2. Cover one of the beakers with a bell jar. 

3. Leave the beakers and, over the course of a day or two, observe how the water level in the two beakers 
changes. What do you notice? Note: You could speed up this demonstration by placing the two 
beakers over a bunsen burner to heat the water. In this case, it may be easier to cover the second 
beaker with a glass cover. 



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65 

Observations: 

You should notice that in the beaker that is uncovered, the water level drops quickly because of evapora- 
tion. In the beaker that is covered, there is an initial drop in the water level, but after a while evaporation 
appears to stop and the water level in this beaker is higher than that in the one that is open. Note that the 
diagram below shows the situation ate time=0. 



Image notjtnished 

Figure 3.10 



Discussion: 

In the first beaker, liquid water becomes water vapour as a result of evaporation and the water level 
drops. In the second beaker, evaporation also takes place. However, in this case, the vapour comes into 
contact with the surface of the bell jar and it cools and condenses to form liquid water again. This water is 
returned to the beaker. Once condensation has begun, the rate at which water is lost from the beaker will 
start to decrease. At some point, the rate of evaporation will be equal to the rate of condensation above the 
beaker, and there will be no change in the water level in the beaker. This can be represented as follows: 

liquid ^^ vapour 

In this example, the reaction (in this case, a change in the phase of water) can proceed in either direction. 
In one direction there is a change in phase from liquid to vapour. But the reverse can also take place, when 
vapour condenses to form water again. 

In a closed system it is possible for reactions to be reversible, such as in the demonstration above. 
In a closed system, it is also possible for a chemical reaction to reach equilibrium. We will discuss these 
concepts in more detail. 

3.3.1.2 Open and closed systems 

An open system is one in which matter or energy can flow into or out of the system. In the liquid- vapour 
demonstration we used, the first beaker was an example of an open system because the beaker could be 
heated or cooled (a change in energy), and water vapour (the matter) could evaporate from the beaker. 

A closed system is one in which energy can enter or leave, but matter cannot. The second beaker 
covered by the bell jar is an example of a closed system. The beaker can still be heated or cooled, but water 
vapour cannot leave the system because the bell jar is a barrier. Condensation changes the vapour to liquid 
and returns it to the beaker. In other words, there is no loss of matter from the system. 

Definition 3.5: Open and closed systems 

An open system is one whose borders allow the movement of energy and matter into and out of 
the system. A closed system is one in which only energy can be exchanged, but not matter. 

3.3.1.3 Reversible reactions 

Some reactions can take place in two directions. In one direction the reactants combine to form the products. 
This is called the forward reaction. In the other, the products react to form reactants again. This is called 
the reverse reaction. A special double-headed arrow is used to show this type of reversible reaction: 

XY + Z^X + YZ 

So, in the following reversible reaction: 

H2{g)+l2ig)-2HI{g) 

The forward reaction is H2 {g) + I2 (9) -^ 2HI {g). The reverse reaction is 2HI {g) -^ H2 {g) + I2 [g)- 



66 CHAPTER 3. REACTION RATES 

Definition 3.6: A reversible reaction 

A reversible reaction is a chemical reaction that can proceed in both the forward and reverse 
directions. In other words, the reactant and product of one reaction may reverse roles. 

3.3.1.3.1 Demonstration : The reversibility of chemical reactions 

Apparatus and materials: 

Lime water (Ca(0H)2); calcium carbonate (CaCOs); hydrochloric acid; 2 test tubes with rubber stoppers; 
delivery tube; retort stand and clamp; bunsen burner. 
Method and observations: 

1. Half-fill a test tube with clear lime water (Ca(0H)2). 

2. In another test tube, place a few pieces of calcium carbonate (CaCOs) and cover the pieces with dilute 
hydrochloric acid. Seal the test tube with a rubber stopper and delivery tube. 

3. Place the other end of the delivery tube into the test tube containing the lime water so that the carbon 
dioxide that is produced from the reaction between calcium carbonate and hydrochloric acid passes 
through the lime water. Observe what happens to the appearance of the lime water. The equation for 
the reaction that takes place is: Ca{OH)^ + CO2 -^ CaCO^ + H2O CaCOa is insoluble and it turns 
the limewater milky. 

4. Allow the reaction to proceed for a while so that carbon dioxide continues to pass through the limewater. 
What do you notice? The equation for the reaction that takes place is: CaCO^ (s) + H2O + CO2 -^ 
Ca{HC0y,)2 III this reaction, calcium carbonate becomes one of the reactants to produce hydrogen 
carbonate (Ca(HC03)2) and so the solution becomes clear again. 

5. Heat the solution in the test tube over a bunsen burner. What do you observe? You should see bubbles 
of carbon dioxide appear and the limewater turns milky again. The reaction that has taken place is: 
Ca{HC03)2 -^ CaCOs (s) + H2O + CO2 



Image notjinished 



Figure 3.11 



Discussion: 

• If you look at the last two equations you will see that the one is the reverse of the other. In other words, 
this is a reversible reaction and can be written as follows: CaCOs (s) + H2O + CO2 ^ Ca{HCO'i)2 

• Is the forward reaction endothermic or exothermic? Is the reverse reaction endothermic or exothermic? 
You should have noticed that the reverse reaction only took place when the solution was heated. 
Sometimes, changing the temperature of a reaction can change its direction. 



3.3.1.4 Chemical equilibrium 

Using the same reversible reaction that we used in an earlier example: 
H2{g)+l2{g)-2HI{g) 
The forward reaction is: 



H2 + l2^ 2HI (3.6) 



67 

The reverse reaction is: 

2HI ^H^ + h (3.7) 

When the rate of the forward reaction and the reverse reaction are equal, the system is said to be in 
equilbrium. Figure 3.12 shows this. Initially (time = 0), the rate of the forward reaction is high and the 
rate of the reverse reaction is low. As the reaction proceeds, the rate of the forward reaction decreases and 
the rate of the reverse reaction increases, until both occur at the same rate. This is called equilibrium. 

Although it is not always possible to observe any macroscopic changes, this does not mean that the 
reaction has stopped. The forward and reverse reactions continue to take place and so microscopic changes 
still occur in the system. This state is called dynamic equilibrium. In the liquid- vapour phase equilibrium 
demonstration, dynamic equilibrium was reached when there was no observable change in the level of the 
water in the second beaker even though evaporation and condensation continued to take place. 

Image notjinished 

Figure 3.12: The change in rate of forward and reverse reactions in a closed system 



There are, however, a number of factors that can change the chemical equilibrium of a reaction. Changing 
the concentration, the temperature or the pressure of a reaction can affect equilibrium. These factors 
will be discussed in more detail later in this chapter. 

Definition 3.7: Chemical equilibrium 

Chemical equilibrium is the state of a chemical reaction, where the concentrations of the reactants 
and products have no net change over time. Usually, this state results when the forward chemical 
reactions proceed at the same rate as their reverse reactions. 

3.4 The equilibrium constant' 
3.4,1 The equilibrium constant 

Definition 3.8: Equilibrium constant 

The equilibrium constant (Kc), relates to a chemical reaction at equilibrium. It can be calculated 
if the equilibrium concentration of each reactant and product in a reaction at equilibrium is known. 

3.4.1.1 Calculating the equilibrium constant 

Consider the following generalised reaction which takes place in a closed container at a constant temper- 
ature: 

A+ B ^C + D 

We know from "Factors affecting reaction rates" that the rate of the forward reaction is directly propor- 
tional to the concentration of the reactants. In other words, as the concentration of the reactants increases, 
so does the rate of the forward reaction. This can be shown using the following equation: 

Rate of forward reaction ex [A] [B] 

or 

Rate of forward reaction = ki[A][B] 



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68 CHAPTER 3. REACTION RATES 

Similarly, the rate of the reverse reaction is directly proportional to the concentration of the products. 
This can be shown using the following equation: 

Rate of reverse reaction ex [C][D] 

or 

Rate of reverse reaction = k2[C][D] 

At equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction. This can be 
shown using the following equation: 

fci [A] [B] = h [C] [D] (3.8) 

or 

fci [C] [D] 



fc2 [A] [B] 
or, if the constants ki and k2 are simplified to a single constant, the equation becomes: 



(3.9) 



fee - j^j j^j (3.1U) 

A more general form of the equation for a reaction at chemical equilibrium is: 
aA + bB^cC+ dD 

where A and B are reactants, C and D are products and a, b, c, and d are the coefficients of the respective 
reactants and products. A more general formula for calculating the equilibrium constant is therefore: 

A-.^HH! ,3.n) 

[AflB]'' 

It is important to note that if a reactant or a product in a chemical reaction is in either the liquid or solid 
phase, the concentration stays constant during the reaction. Therefore, these values can be left out of the 
equation to calculate Kc. For example, in the following reaction: 
C{s) + H20{g)^CO{g) + H2{g) 

K. = raS (3.12) 

[H2O] ^ ' 

TIP: 

l.The constant Kc is affected by temperature and so, if the values of K^ are being compared 
for different reactions, it is important that all the reactions have taken place at the same 
temperature. 

2.Kc values do not have units. If you look at the equation, the units all cancel each other out. 

3.4.1.2 The meaning of K^ values 

The formula for Kc has the concentration of the products in the numerator and the concentration of reactants 
in the denominator. So a high Kc value means that the concentration of products is high and the reaction 
has a high yield. We can also say that the equilibrium lies far to the right. The opposite is true for a low 
Kc value. A low Kc value means that, at equilibrium, there are more reactants than products and therefore 
the yield is low. The equilibrium for the reaction lies far to the left. 

TIP: Calculations made easy 

When you are busy with calculations that involve the equilibrium constant, the following tips may help: 



69 



1. Make sure that you always read the question carefully to be sure of what you are being asked to 
calculate. If the equilibrium constant is involved, make sure that the concentrations you use are the 
concentrations at equilibrium, and not the concentrations or quantities that are present at some 
other time in the reaction. 

2. When you are doing more complicated calculations, it sometimes helps to draw up a table like the one 
below and fill in the mole values that you know or those you can calculate. This will give you a clear 
picture of what is happening in the reaction and will make sure that you use the right values in your 
calculations. 





Reactant 1 


Reactant 2 


Product 1 


Start of reaction 








Used up 








Produced 








Equilibrium 









Table 3.4 



Exercise 3.2: Calculating Kc 

For the reaction: 

SO2 {g) + NO2 (5) ^ NO [g) + ^03 (5) 

the concentration of the reagents is as follows: 

[SO3] = 0.2 mol.dm-3 

[NO2] = 0.1 mol.dm-3 

[NO] = 0.4 mol.dm-3 

[SO2] = 0.2 mol.dm-3 

Calculate the value of Kc. 

Exercise 3.3: Calculating reagent concentration 

For the reaction: 

S{s) + 02{g)-S02{g) 



(Solution on p. 78.) 



(Solution on p. 78.) 



1. Write an equation for the equilibrium constant. 

2. Calculate the equilibrium concentration of O2 if Kc=6 and [SO2 



'imol.dm ^ at equilibrium. 



Exercise 3.4: Equilibrium calculations (Solution on p. 78.) 

Initially 1.4 moles of NH3(g) is introduced into a sealed 2.0 dm~^ reaction vessel. The ammonia 
decomposes when the temperature is increased to 600K and reaches equilibrium as follows: 

2NH^ {g) ^ N2 (5) + 3^2 {g) 

When the equilibrium mixture is analysed, the concentration of NH3(g) is 0.3 mol-dm~^ 

1. Calculate the concentration of N2(g) and H2(g) in the equilibrium mixture. 

2. Calculate the equilibrium constant for the reaction at 900 K. 



Exercise 3.5: Calculating Kj, (Solution on p. 79.) 

Hydrogen and iodine gas react according to the following equation: 

H2{g) + l2{g)-2HI{g) 

When 0.496 mol H2 and 0.181 mol I2 are heated at 450°C in a 1 dm"^ container, the equilibrium 
mixture is found to contain 0.00749 mol 12- Calculate the equilibrium constant for the reaction at 
450°C. 



70 CHAPTER 3. REACTION RATES 

3.4.1.2.1 The equilibrium constant 

1. Write the equiUbrium constant expression, K^ for the following reactions: 

a. 2N0 (g) + Ch (<?) ^ 2N0C1 

2. The following reaction takes place: Fe^"*" (aq) + 4Cn ^^ FeCl^ (aq) Kc for the reaction is 7.5 x 10^^ 
moLdm""^. At equilibrium, the concentration of FeCl^ is 0.95 x 10""* moLdm""^ and the concentration 
of free iron (Fe^+) is 0.2 moLdm""^. Calculate the concentration of chloride ions at equilibrium. 

3. Ethanoic acid (CH3COOH) reacts with ethanol (CH3CH2OH) to produce ethyl ethanoate and water. 
The reaction is: CH3COOH + CH3CH2OH -^ CH3COOCH2CH3 + H2O At the beginning of the 
reaction, there are 0.5 mols of ethanoic acid and 0.5 mols of ethanol. At equilibrium, 0.3 mols of 
ethanoic acid was left unreacted. The volume of the reaction container is 2 dm^. Calculate the value 
of K,. 



3.5 Le Chateliers principle'' 

3.5.1 Le Chatelier's principle 

A number of factors can influence the equilibrium of a reaction. These are: 

1. concentration 

2. temperature 

3. pressure 

Le Chatelier's Principle helps to predict what a change in temperature, concentration or pressure will 
have on the position of the equilibrium in a chemical reaction. This is very important, particularly in 
industrial applications, where yields must be accurately predicted and maximised. 

Definition 3.9: Le Chatelier's Principle 

If a chemical system at equilibrium experiences a change in concentration, temperature or total 
pressure the equilibrium will shift in order to minimise that change and a new equilibrium is 
established. 

3.5.1.1 The effect of concentration on equilibrium 

If the concentration of a substance is increased, the equilibrium will shift so that this concentration decreases. 
So for example, if the concentration of a reactant was increased, the equilibrium would shift in the direction 
of the reaction that uses up the reactants, so that the reactant concentration decreases and equilibrium is 
restored. In the reaction between nitrogen and hydrogen to produce ammonia: 

N2 {g) + 3^2 {g) - 2NH3 (g) 

• If the nitrogen or hydrogen concentration was increased, Le Chatelier's principle predicts that equilib- 
rium will shift to favour the forward reaction so that the excess nitrogen and hydrogen are used up to 
produce ammonia. Equilibrium shifts to the right. 

• If the nitrogen or hydrogen concentration was decreased, the reverse reaction would be favoured so 
that some of the ammonia would change back to nitrogen and hydrogen to restore equilibrium. 

• The same would be true if the concentration of the product (NH3) was changed. If [NH3] decreases, 
the forward reaction is favoured and if [NH3] increases, the reverse reaction is favoured. 



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71 

3.5.1.2 The effect of temperature on equilibrium 

If the temperature of a reaction mixture is increased, the equihbrium will shift to decrease the temperature. 
So it will favour the reaction which will use up heat energy, in other words the endothermic reaction. The 
opposite is true if the temperature is decreased. In this case, the reaction that produces heat energy will be 
favoured, in other words, the exothermic reaction. 

The reaction shown below is exothermic (shown by the negative value for A H). This means that the 
forward reaction, where nitrogen and hydrogen react to form ammonia, gives off heat. In the reverse reaction, 
where ammonia is broken down into hydrogen and nitrogen gas, heat is used up and so this reaction is 
endothermic. 

e.g. N2 (g) + 3H2 (g) ^ 2NH3 (g) and AH = -92kJ (3.13) 

An increase in temperature favours the reaction that is endothermic (the reverse reaction) because it uses 
up energy. If the temperature is increased, then the yield of ammonia (NH3) decreases. 

A decrease in temperature favours the reaction that is exothermic (the forward reaction) because it 
produces energy. Therefore, if the temperature is decreased, then the yield of NH3 increases. 

Khan academy video on Le Chateliers principle 

This media object is a Flash object. Please view or download it at 
<http://www.youtube.com/v/4-fEvpVNTlE&rel=0&hl=en_US&feature=player_embedded&version=3> 

Figure 3.13 



3.5.1.2.1 Experiment : Le Chatelier's Principle 

Aim: 

To determine the effect of a change in concentration and temperature on chemical equilibrium 

Apparatus: 

0.2 M C0CI2 solution, concentrated HCl, water, test tube, bunsen burner 

Method: 

1. Put 4-5 drops of 0.2M C0CI2 solution into a test tube. 

2. Add 20-25 drops of concentrated HCl. 

3. Add 10-12 drops of water. 

4. Heat the solution for 1-2 minutes. 

5. Cool the solution for 1 minute under a tap. 

6. Observe and record the colour changes that take place during the reaction. 

The equation for the reaction that takes place is: 

e.g. CoCll-+6H20^CoiH20)l++ACl- ,_.. 

^ V ' ^ V ' W-J^^J 

blue pink 

Results: 

Complete your observations in the table below, showing the colour changes that take place, and also 
indicating whether the concentration of each of the ions in solution increases or decreases. 



72 



CHAPTER 3. REACTION RATES 





Initial colour 


Final colour 


[Co^+] 


[CI ] 


[CociM 


Add Cr 












Add H2O 












Increase temp. 












Decrease temp. 













Table 3.5 

Conclusions: 

Use your knowledge of equilibrium principles to explain the changes that you recorded in the table above. 
Draw a conclusion about the effect of a change in concentration of either the reactants or products on the 
equilibrium position. Also draw a conclusion about the effect of a change in temperature on the equilibrium 
position. 

3.5.1.3 The effect of pressure on equilibrium 

In the case of gases, we refer to pressure instead of concentration. Similar principles apply as those that 
were described before for concentration. When the pressure of a system increases, there are more particles 
in a particular space. The equilibrium will shift in a direction that reduces the number of gas particles so 
that the pressure is also reduced. To predict what will happen in a reaction, we need to look at the number 
of moles of gas that are in the reactants and products. Look at the example below: 



e.g. 2SO2 {9) + O2 (g) - 2SO3 (g) 



(3.15) 



In this reaction, two moles of product are formed for every three moles of reactants. If we increase the 
pressure on the closed system, the equilibrium will shift to the right because the forward reaction reduces the 
number of moles of gas that are present. This means that the yield of SO3 will increase. The opposite will 
apply if the pressure on the system decreases, the equilibrium will shift to the left, and the concentration of 
SO2 and O2 will increase. 



TIP: 



1. If the forward reaction that forms the product is endothermic, then an increase in temperature 

will favour this reaction and the yield of product will increase. Lowering the temperature will 

decrease the product yield. 
2. If the forward reaction that forms the product is exothermic, then a decrease in temperature 

will favour this reaction and the product yield will increase. Increasing the temperature will 

decrease the product yield. 
3. Increasing the pressure favours the side of the equilibrium with the least number of gas 

molecules. This is shown in the balanced symbol equation. This rule applies in reactions 

with one or more gaseous reactants or products. 
4. Decreasing the pressure favours the side of the equilibrium with the most number of gas 

molecules. This rule applies in reactions with one or more gaseous reactants or products. 
5. If the concentration of a reactant (on the left) is increased, then some of it must change to the 

products (on the right) for equilibrium to be maintained. The equilibrium position will shift 

to the right. 
6. If the concentration of a reactant (on the left) is decreased, then some of the products (on 

the right) must change back to reactants for equilibrium to be maintained. The equilibrium 

position will shift to the left. 
7.A catalyst does not affect the equilibrium position of a reaction. It only influences the rate 

of the reaction, in other words, how quickly equilibrium is reached. 



73 



Exercise 3.6: Reaction Rates 1 (Solution on p. 79.) 

2NO2 (5) ^ 2NO {g) + O2 {g) and AH > How will the rate of the reverse reaction be affected 
by: 

1. a decrease in temperature? 

2. the addition of a catalyst? 

3. the addition of more NO gas? 



Exercise 3.7: Reaction Rates 2 



(Solution on p. 80.) 



1. Write a balanced equation for the exothermic reaction between Zn(s) and HCl. 

2. Name 3 ways to increase the reaction rate between hydrochloric acid and zinc metal. 



3.5.1.3.1 Reaction rates and equilibrium 

1. The following reaction reaches equilibrium in a closed container: CaCOs (s) ^^ CaO {s) + C02 (g) The 
pressure of the system is increased by decreasing the volume of the container. How will the number 
of moles and the concentration of the C02(g) have changed when a new equilibrium is reached at the 
same temperature? 





moles of CO2 


[CO2] 


A 


decreased 


decreased 


B 


increased 


increased 


C 


decreased 


stays the same 


D 


decreased 


increased 



Table 3.6 

(lEB Paper 2, 2003) 
2. The following reaction has reached equilibrium in a closed container: C (s) + H2O {g) ^^ CO {g) + 
H2 {g)AH > The pressure of the system is then decreased by increasing the volume of the container. 
How will the concentration of the H2(g) and the value of Kc be affected when the new equilibrium is 
established? Assume that the temperature of the system remains unchanged. 





[H2] 


Ke 


A 


increases 


increases 


B 


increases 


unchanged 


C 


unchanged 


unchanged 


D 


decreases 


unchanged 



Table 3.7 

(lEB Paper 2, 2004) 
3. During a classroom experiment copper metal reacts with concentrated nitric acid to produce NO2 gas, 
which is collected in a gas syringe. When enough gas has collected in the syringe, the delivery tube is 
clamped so that no gas can escape. The brown NO2 gas collected reaches an equilibrium with colourless 
N2O4 gas as represented by the following equation: 



2NO2 (<?) ^ N2O4 (5) 



(3.16) 



74 CHAPTER 3. REACTION RATES 

Once this equilibrium has been established, there are 0.01 moles of NO2 gas and 0.03 moles of N2O4 
gas present in the syringe. 

a. A learner, noticing that the colour of the gas mixture in the syringe is no longer changing, 
comments that all chemical reactions in the syringe must have stopped. Is this assumption 
correct? Explain. 

b. The gas in the syringe is cooled. The volume of the gas is kept constant during the cooling process. 
Will the gas be lighter or darker at the lower temperature? Explain your answer. 

c. The volume of the syringe is now reduced to 75 cm^ by pushing the plunger in and holding it 
in the new position. There are 0.032 moles of N2O4 gas present once the equilibrium has been 
re-established at the reduced volume (75 cm'^). Calculate the value of the equilibrium constant 
for this equilibrium. (lEB Paper 2, 2004) 

4. Consider the following reaction, which takes place in a closed container: A{s) + B {g) -^ AB (5) AH < 
If you wanted to increase the rate of the reaction, which of the following would you do? 

a. decrease the concentration of B 

b. decrease the temperature of A 

c. grind A into a fine powder 

d. decrease the pressure 

(lEB Paper 2, 2002) 

5. Gases X and Y are pumped into a 2 dm^ container. When the container is sealed, 4 moles of gas X 
and 4 moles of gas Y are present. The following equilibrium is established: 2X {g) + SF {g) ^^ ^^2^3 
The graph below shows the number of moles of gas X and gas X2Y3 that are present from the time 
the container is sealed. 



Image notjinished 

Figure 3.14 



a. How many moles of gas X2Y3 are formed by the time the reaction reaches equilibrium at 30 
seconds? 

b. Calculate the value of the equilibrium constant at t = 50 s. 

c. At 70 s the temperature is increased. Is the forward reaction endothermic or exothermic? Explain 
in terms of Le Chatelier's Principle. 

d. How will this increase in temperature affect the value of the equilibrium constant? 

3.5.2 Industrial applications 

The Haber process is a good example of an industrial process which uses the equilibrium principles that 
have been discussed. The equation for the process is as follows: 

N2 (g) + 3H2 (g) ^ 2NH3 (g) + energy (3.17) 

Since the reaction is exothermic, the forward reaction is favoured at low temperatures, and the reverse 
reaction at high temperatures. If the purpose of the Haber process is to produce ammonia, then the temper- 
ature must be maintained at a level that is low enough to ensure that the reaction continues in the forward 
direction. 

The forward reaction is also favoured by high pressures because there are four moles of reactant for every 
two moles of product formed. 



75 



The K value for this reaction will be calculated as follows: 

l2 



K 



[NH: 



3J 



[N2] [H2 



(3.18) 



3.5.2.1 Applying equilibrium principles 

Look at the values of k calculated for the Haber process reaction at different temperatures, and then answer 
the questions that follow: 



j'oC 


k 


25 


6.4 X 10^ 


200 


4.4 X W-^ 


300 


4.3 X 10~^ 


400 


1.6 X 10"'' 


500 


1.5 X 10-5 



Table 3.8 

1. What happens to the value of K as the temperature increases? 

2. Which reaction is being favoured when the temperature is 300 degrees Celsius? 

3. According to this table, which temperature would be best if you wanted to produce as much ammonia 
as possible? Explain. 

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Figure 3.15 



3.5.3 Summary 

• The rate of a reaction describes how quickly reactants are used up, or how quickly products form. 
The units used are moles per second. 

• A number of factors can affect the rate of a reaction. These include the nature of the reactants, 
the concentration of reactants, temperature of the reaction, the presence or absence of a catalyst 
and the surface area of the reactants. 

• Collision theory provides one way of explaining why each of these factors can affect the rate of 
a reaction. For example, higher temperatures mean increased reaction rates because the reactant 
particles have more energy and are more likely to collide successfully with each other. 

• Different methods can be used to measure the rate of a reaction. The method used will depend 
on the nature of the product. Reactions that produce gases can be measured by collecting the gas in a 
syringe. Reactions that produce a precipitate are also easy to measure because the precipitate is easily 
visible. 



76 CHAPTER 3. REACTION RATES 

• For any reaction to occur, a minimum amount of energy is needed so that bonds in the reactants can 
break, and new bonds can form in the products. The minimum energy that is required is called the 
activation energy of a reaction. 

• In reactions where the particles do not have enough energy to overcome this activation energy, one of 
two methods can be used to facilitate a reaction to take place: increase the temperature of the reaction 
or add a catalyst. 

• Increasing the temperature of a reaction means that the average energy of the reactant particles 
increases and they are more likely to have enough energy to overcome the activation energy. 

• A catalyst is used to lower the activation energy so that the reaction is more likely to take place. A 
catalyst does this by providing an alternative, lower energy pathway, for the reaction. 

• A catalyst therefore speeds up a reaction but does not become part of the reaction in any way. 

• Chemical equilibrium is the state of a reaction, where the concentrations of the reactants and the 
products have no net change over time. Usually this occurs when the rate of the forward reaction is 
the same as the rate of the reverse reaction. 

• The equilibrium constant relates to reactions at equilibrium, and can be calculated using the fol- 
lowing equation: 

|A]°|B1» 

where A and B are reactants, C and D are products and a, b, c, and d are the coefficients of the 
respective reactants and products. 

• A high Kc value means that the concentration of products at equilibrium is high and the reaction 
has a high yield. A low Kc value means that the concentration of products at equilibrium is low and 
the reaction has a low yield. 

• Le Chatelier's Principle states that if a chemical system at equilibrium experiences a change in 
concentration, temperature or total pressure the equilibrium will shift in order to minimise that change 
and to re-establish equilibrium. For example, if the pressure of a gaseous system at eqilibrium was 
increased, the equilibrium would shift to favour the reaction that produces the lowest quantity of the 
gas. If the temperature of the same system was to increase, the equilibrium would shift to favour 
the endothermic reaction. Similar principles apply for changes in concentration of the reactants or 
products in a reaction. 

• The principles of equilibrium are very important in industrial applications such as the Haber process, 
so that productivity can be maximised. 

3.5.3.1 Summary Exercise 

1. For each of the following questions, choose the one correct answer from the list provided. 

a. Consider the following reaction that has reached equilibrium after some time in a sealed 1 dm^ 
flask: PCI5 {g) ^^ PCI3 {g)+Cl2 (5); Ai7 is positive Which one of the following reaction conditions 
applied to the system would decrease the rate of the reverse reaction? 

1. increase the pressure 

2. increase the reaction temperature 

3. continually remove Cl2(g) from the flask 

4. addition of a suitable catalyst 
(lEB Paper 2, 2001) 

b. The following equilibrium constant expression is given for a particular reaction: Kc = 
[H2O] [CO2] I [Cs-ffs] [O2] For which one of the following reactions is the above expression of 
Kc is correct? 

1. CsFs (5) + 5O2 (ff) ^ 4i720 (g) + 3CO2 (5) 

2. m^O (3) + 3CO2 (ff) ^ Csilg (5) + 5O2 {g) 

3. 2^3/^8 (5) + 7O2 {g) - QCO {g) + 8H2O (g) 



77 

4. ^3^8 {g) + 502 {g) - 4i720 (0 + 3C02 (s) 
(lEB Paper 2, 2001) 

2. 10 g of magnesium ribbon reacts with a 0.15 mol.dm"'^ solution of hydrochloric acid at a temperature 
of 25OC. 

a. Write a balanced chemical equation for the reaction. 

b. State two ways of increasing the rate of production of H2(g). 

c. A table of the results is given below: 



Time elapsed (min) 


Volof H2(g) (cm3) 








0.5 


17 


1.0 


25 


1.5 


30 


2.0 


33 


2.5 


35 


3.0 


35 



Table 3.9 

1. Plot a graph of volume versus time for these results. 

2. Explain the shape of the graph during the following two time intervals: t = to t = 2.0 min 
and then t = 2.5 and t = 3.0 min by referring to the volume of H2(g) produced. 

(lEB Paper 2, 2001) 
3. Cobalt chloride crystals are dissolved in a beaker containing ethanol and then a few drops of water are 
added. After a period of time, the reaction reaches equilibrium as follows: CoCl^~ (blue) +6H2O ^^ 
Co {H20)q (pink) +ACl~ The solution, which is now just blue, is poured into three test tubes. State, 
in each case, what colour changes will be observed (if any) if the following are added in turn to each 
test tube: 

a. 1 cm^ of distilled water 

b. A few crystals of sodium chloride 

c. The addition of dilute hydrochloric acid to the third test tube causes the solution to turn pink. 
Explain why this occurs. 

(lEB Paper 2, 2001) 



78 CHAPTER 3. REACTION RATES 

Solutions to Exercises in Chapter 3 

Solution to Exercise 3.1 (p. 54) 

Step 1. 

TTl 4 

n= — = = 0.58mo/s (3.20) 

M 6.94 ^ ' 



Step 2. 

Step 3. Rate of reaction = 



t = 2 X 60 = 120s (3.21) 

moles of Lithium used 0.58 



time 120 

The rate of the reaction is 0.005 mol.s"^ 

Solution to Exercise 3.2 (p. 69) 

Step 1. 



0.005 (3.22) 



^'-[SO,][NO,] ^^-^^^ 



i^c=^i^=4 (3.24) 

(0.2x0.1) ^ ^ 



Step 2. 

Solution to Exercise 3.3 (p. 69) 

Step 1. 

(Sulfur is left out of the equation because it is a solid and its concentration stays constant during the 
reaction) 
Step 2. 

[O.] = ^ (3.26) 



Step 3. 

[O2] = = 0.5mo/.rfm"^ (3.27) 

Solution to Exercise 3.4 (p. 69) 

Step 1. 

ft 

c = - (3.28) 

Therefore, 

n = cxT/ = 0.3x2 = 0.6mol (3.29) 

Step 2. Moles used up = 1.4 - 0.6 = 0.8 moles 

Step 3. Remember to use the mole ratio of reactants to products to do this. In this case, the ratio of NH3:N2:H2 

= 2:1:3. Therefore, if 0.8 moles of ammonia are used up in the reaction, then 0.4 moles of nitrogen are 

produced and 1.2 moles of hydrogen are produced. 



79 



Step 5. 



Step 4. 





NH3 


N2 


H2 


Start of reaction 


1.4 








Used up 


0.8 








Produced 





0.4 


1.2 


Equilibrium 


0.6 


0.4 


1.2 



Table 3.10 



[H2] 



n 
V 

n 

V 



0.4 
1.2 



0.2 mol.dm '^ 
0.6 mol.dm~ 



Step 6. 



Solution to Exercise 3.5 (p. 69) 



Kr 



[H2f [N2] (0.6)' (0.2) 



[NH3 



my 



0.48 



(3.30) 
(3.31) 

(3.32) 



Step 1. Moles of iodine used = 0.181 - 0.00749 = 0.1735 mol 

Step 2. The mole ratio of hydrogendodine = 1:1, therefore 0.1735 moles of hydrogen must also be used up in 

the reaction. 
Step 3. The mole ratio of H2:l2:HI = 1:1:2, therefore the number of moles of HI produced is 0.1735 x 2 = 

0.347 mol. 

So far, the table can be filled in as follows: 





H2 (g) 


I2 


2HI 


Start of reaction 


0.496 


0.181 





Used up 


0.1735 


0.1735 





Produced 








0.347 


Equilibrium 


0.3225 


0.0075 


0.347 



Step 4. 



Table 3.11 



V 



Step 5. 



Therefore the equilibrium concentrations are as follows: 
[H2] = 0.3225 mol.dm-3 
[I2] = 0.0075 mol.dm-3 
[HI] = 0.347 mol.dm-3 



K, 



[HI] 
{H2] [I2 



0.347 



0.3225 X 0.0075 



143.47 



(3.33) 



(3.34) 



Solution to Exercise 3.6 (p. 73) 



80 CHAPTER 3. REACTION RATES 

Step 1. The rate of the forward reaction will increase since it is the forward reaction that is exothermix and 
therefore produces energy to balance the loss of energy from the decrease in temperature. The rate of 
the reverse reaction will decrease. 

Step 2. The rate of the reverse and the forward reaction will increase. 

Step 3. The rate of the reverse reaction will increase so that the extra NO gas is converted into NO2 gas. 

Solution to Exercise 3.7 (p. 73) 

Step 1. Zn (s) + 2HCI {aq) ^ ZnCh {aq) + H2 (g) 

Step 2. A catalyst could be added, the zinc solid could be ground into a fine powder to increase its surface 
area, the HCl concentration could be increased or the reaction temperature could be increased. 



Chapter 4 

Electrochemical reactions 



4.1 The galvanic cell' 

4.1.1 Introduction 

Reaction Types in Grade 11 discussed oxidation, reduction and redox reactions. Oxidation involves a loss 
of electrons and reduction involves a gain of electrons. A redox reaction is a reaction where both 
oxidation and reduction take place. What is common to all of these processes is that they involve a transfer 
of electrons and a change in the oxidation state of the elements that are involved. 

4.1.1.1 Oxidation and reduction 

1. Define the terms oxidation and reduction. 

2. In each of the following reactions say whether the iron in the reactants is oxidised or reduced. 

a. Fe -^ Fe'^+ + 2e- 

b. Fe^+ + e- ^ Fe^+ 

c. FeaOs + 3C0 -^ 2Fe + SCOa 

d. Fe^+ -^ Fe^+ + e" 

e. FeaOs + 2AI -^ AI2O3 + 2Fe 

3. In each of the following equations, say which elements in the reactants are oxidised and which are 
reduced. 

a. CuO (s) + H2 [g] -^ Cu [s] + H2O [g] 

b. 2NO {g) + 2C0 {g) -^ N^ {g) + 2CO2 (5) 

c. Mg (s) + FeSOi {aq) -^ MgSOi {aq) + Fe (s) 

d. Zn (s) + 2AgN03 {aq) -^ 2Ag + ZniNOs)^ {aq) 

4. Which one of the substances listed below acts as the oxidising agent in the following reaction? 35'02 + 
Cr20'^- + 2H+ -^ ^SOl' + 2Cr^+ + H2O 

a. H+ 

b. Cr3+ 

c. SO2 

d. CraO?- 

In Grade 11, an experiment was carried out to see what happened when zinc granules are added to a solution 
of copper(II) sulphate. In the experiment, the Cu^"*" ions from the copper(II) sulphate solution were reduced 



^This content is available online at <http://siyavula.cnx.Org/content/m39417/l.l/>. 



81 



82 CHAPTER 4. ELECTROCHEMICAL REACTIONS 

to copper metal, which was then deposited in a layer on the zinc granules. The zinc atoms were oxidised to 
form Zn^"*" ions in the solution. The half reactions are as follows: 

Cu^+ {aq) + 2e^ -^ Cu (s) (reduction half reaction) 

Zn (s) -^ Zn?^ (aq) + 2e^ (oxidation half reaction) 

The overall redox reaction is: 

Cu^+ {aq) + Zn^Cu (s) + Zn^+ {aq) 

There was an increase in the temperature of the reaction when you carried out this experiment. Is it 
possible that this heat energy could be converted into electrical energy? In other words, can we use a chemical 
reaction where there is an exchange of electrons, to produce electricity? And if this is possible, what would 
happen if an electrical current was supplied to cause some type of chemical reaction to take place? 

An electrochemical reaction is a chemical reaction that produces a voltage, and therefore a flow of 
electrical current. An electrochemical reaction can also be the reverse of this process, in other words if an 
electrical current causes a chemical reaction to take place. 

Definition 4.1: Electrochemical reaction 

If a chemical reaction is caused by an external voltage, or if a voltage is caused by a chemical 
reaction, it is an electrochemical reaction. 

Electrochemistry is the branch of chemistry that studies these electrochemical reactions. In this 
chapter, we will be looking more closely at different types of electrochemical reactions, and how these can 
be used in different ways. 

4.1.2 The Galvanic Cell 

4.1.2.1 Experiment : Electrochemical reactions 

Aim: 

To investigate the reactions that take place in a zinc-copper cell 

Apparatus: 

zinc plate, copper plate, measuring balance, zinc sulphate (ZnS04) solution (1 mol.dm"^), copper sul- 
phate (CUSO4) solution (1 mol.dm"^), two 250 ml beakers, U-tube, Na2S04 solution, cotton wool, ammeter, 
connecting wire. 

Method: 

1. Measure the mass of the copper and zinc plates and record your findings. 

2. Pour about 200 ml of the zinc sulphate solution into a beaker and put the zinc plate into it. 

3. Pour about 200 ml of the copper sulphate solution into the second beaker and place the copper plate 
into it. 

4. Fill the U-tube with the Na2S04 solution and seal the ends of the tubes with the cotton wool. This 
will stop the solution from flowing out when the U-tube is turned upside down. 

5. Connect the zinc and copper plates to the ammeter and observe whether the ammeter records a reading. 

6. Place the U-tube so that one end is in the copper sulphate solution and the other end is in the zinc 
sulphate solution. Is there a reading on the ammeter? In which direction is the current flowing? 

7. Take the ammeter away and connect the copper and zinc plates to each other directly using copper 
wire. Leave to stand for about one day. 

8. After a day, remove the two plates and rinse them first with distilled water, then with alcohol and 
finally with ether. Dry the plates using a hair dryer. 

9. Weigh the zinc and copper plates and record their mass. Has the mass of the plates changed from the 
original measurements? 

Note: A voltmeter can also be used in place of the ammeter. A voltmeter will measure the potential 
difference across the cell. 



83 



electron flow 




CuS04(aq) 



ZnS04(aq) 



Figure 4.1 



Results: 

During the experiment, you should have noticed the following: 



• When the U-tube containing the Na2S04 solution was absent, there was no reading on the ammeter. 

• When the U-tube was connected, a reading was recorded on the ammeter. 



84 CHAPTER 4. ELECTROCHEMICAL REACTIONS 

• After the plates had been connected directly to each other and left for a day, there was a change in 
their mass. The mass of the zinc plate decreased, while the mass of the copper plate increased. 

• The direction of electron flow is from the zinc plate towards the copper plate. 

Conclusions: 

When a zinc sulphate solution containing a zinc plate is connected by a U-tube to a copper sulphate 
solution containing a copper plate, reactions occur in both solutions. The decrease in mass of the zinc plate 
suggests that the zinc metal has been oxidised. The increase in mass of the copper plate suggests that 
reduction has occurred here to produce more copper metal. This will be explained in detail below. 

4.1.2.2 Half-cell reactions in the Zn-Cu cell 

The experiment above demonstrated a zinc-copper cell. This was made up of a zinc half cell and a copper 
half cell. 

Definition 4.2: Half cell 

A half cell is a structure that consists of a conductive electrode surrounded by a conductive 
electrolyte. For example, a zinc half cell could consist of a zinc metal plate (the electrode) in a zinc 
sulphate solution (the electrolyte). 

How do we explain what has just been observed in the zinc-copper cell? 

• Copper plate At the copper plate, there was an increase in mass. This means that Cu^"*" ions from the 
copper sulphate solution were deposited onto the plate as atoms of copper metal. The half-reaction that 
takes place at the copper plate is: Cu^+ -I- 2e~ -^ CM(Reduction half reaction) Another shortened 
way to represent this copper half-cell is Cu^+ZCu. 

• Zinc plate At the zinc plate, there was a decrease in mass. This means that some of the zinc goes 
into solution as Z^+ ions. The electrons remain on the zinc plate, giving it a negative charge. The 
half-reaction that takes place at the zinc plate is: Zn -^ Ztt?^ + 2e~ (Oxidation half reaction) The 
shortened way to represent the zinc half-cell is Zn/Zn^+. The overall reaction is: Zn + Cu^^ +2e~ -^ 
Zn^'^ + Cu + 2e~ or, if we cancel the electrons: Zn + Cu^^ -^ Zr?^ -I- Cu For this electrochemical 
cell, the standard notation is: 

Zn|Zn2+||Cu2+|CM (4.1) 

where 

I = a phase boundary (solid/aqueous) 

II = the salt bridge 

In the notation used above, the oxidation half-reaction at the anode is written on the left, and the reduction 
half-reaction at the cathode is written on the right. In the Zn-Cu electrochemical cell, the direction of current 
flow in the external circuit is from the zinc electrode (where there has been a build up of electrons) to the 
copper electrode. 

4.1.2.3 Components of the Zn-Cu cell 

In the zinc-copper cell, the copper and zinc plates are called the electrodes. The electrode where oxidation 
occurs is called the anode, and the electrode where reduction takes place is called the cathode. In the 
zinc-copper cell, the zinc plate is the anode and the copper plate is the cathode. 

Definition 4.3: Electrode 

An electrode is an electrical conductor that is used to make contact with a metallic part of a 
circuit. The anode is the electrode where oxidation takes place. The cathode is the electrode where 
reduction takes place. 

The zinc sulphate and copper sulphate solutions are called the electrolyte solutions. 



85 

Definition 4.4: Electrolyte 

An electrolyte is a substance that contains free ions and which therefore behaves as an electrical 
conductor. 

The U-tube also plays a very important role in the cell. In the Zn/Zn^+ half-cell, there is a build up 
of positive charge because of the release of electrons through oxidation. In the Cu^^/Cu half-cell, there is 
a decrease in the positive charge because electrons are gained through reduction. This causes a movement 
of S04~ ions into the beaker where there are too many positive ions, in order to neutralise the solution. 
Without this, the flow of electrons in the outer circuit stops completely. The U-tube is called the salt 
bridge. The salt bridge acts as a transfer medium that allows ions to flow through without allowing the 
different solutions to mix and react. 

Definition 4.5: Salt bridge 

A salt bridge, in electrochemistry, is a laboratory device that is used to connect the oxidation and 
reduction half-cells of a galvanic cell. 

4.1.2.4 The Galvanic cell 

In the zinc-copper cell the important thing to notice is that the chemical reactions that take place at the two 
electrodes cause an electric current to flow through the outer circuit. In this type of cell, chemical energy 
is converted to electrical energy. These are called galvanic cells. The zinc-copper cell is one example 
of a galvanic cell. A galvanic cell (which is also sometimes referred to as a voltaic or electrochemical cell) 
consists of two metals that are connected by a salt bridge between the individual half-cells. A galvanic cell 
generates electricity using the reactions that take place at these two metals, each of which has a different 
reaction potential. 

So what is meant by the 'reaction potential' of a substance? Every metal has a different half reaction and 
different dissolving rates. When two metals with different reaction potentials are used in a galvanic cell, a 
potential difference is set up between the two electrodes, and the result is a flow of current through the wire 
that connects the electrodes. In the zinc-copper cell, zinc has a higher reaction potential than copper and 
therefore dissolves more readily into solution. The metal 'dissolves' when it loses electrons to form positive 
metal ions. These electrons are then transferred through the connecting wire in the outer circuit. 

Definition 4.6: Galvanic cell 

A galvanic (voltaic) cell is an electrochemical cell that uses a chemical reaction between two 
dissimilar electrodes dipped in an electrolyte, to generate an electric current. 

NOTE: It was the Italian physician and anatomist Luigi Galvani who marked the birth of electro- 
chemistry by making a link between chemical reactions and electricity. In 1780, Galvani discovered 
that when two different metals (copper and zinc for example) were connected together and then both 
touched to different parts of a nerve of a frog leg at the same time, they made the leg contract. 
He called this "animal electricity". While many scientists accepted his ideas, another scientist, 
Alessandro Volta, did not. In 1800, because of his professional disagreement over the galvanic re- 
sponse that had been suggested by Luigi Galvani, Volta developed the voltaic pile, which was very 
similar to the galvanic cell. It was the work of these two men that paved the way for all electrical 
batteries. 

Exercise 4.1: Understanding galvanic cells (Solution on p. 103.) 

For the following cell: 

Zn\Zn^+\\Ag+\Ag (4.3) 



1. Give the anode and cathode half- reactions. 

2. Write the overall equation for the chemical reaction. 



86 CHAPTER 4. ELECTROCHEMICAL REACTIONS 

3. Give the direction of the current in the external circuit. 

4.1.2.5 Uses and applications of the galvanic cell 

The principles of the galvanic cell are used to make electrical batteries. In science and technology, a 
battery is a device that stores chemical energy and makes it available in an electrical form. Batteries are 
made of electrochemical devices such as one or more galvanic cells, fuel cells or flow cells. Batteries have 
many uses including in torches, electrical appliances (long-life alkaline batteries), digital cameras (lithium 
batteries), hearing aids (silver-oxide batteries), digital watches (mercury batteries) and military applications 
(thermal batteries). Refer to chapter for more information on batteries. 

The galvanic cell can also be used for electroplating. Electroplating occurs when an electrically con- 
ductive object is coated with a layer of metal using electrical current. Sometimes, electroplating is used to 
give a metal particular properties such as corrosion protection or wear resistance. At other times, it can be 
for aesthetic reasons for example in the production of jewellery. This will be discussed in more detail later 
in this chapter. 

4.1.2.5.1 Galvanic cells 

1. The following half- reactions take place in an electrochemical cell: Fe -^ Fe^^ + 3e~Fe^^ -I- 2e~ -^ Fe 

a. Which is the oxidation half-reaction? 

b. Which is the reduction half-reaction? 

c. Name one oxidising agent. 

d. Name one reducing agent. 

e. Use standard notation to represent this electrochemical cell. 

2. For the following cell: 

Mg\Mg^+\\Mn^+\Mn (4.4) 

a. Give the cathode half-reaction. 

b. Give the anode half-reaction. 

c. Give the overall equation for the electrochemical cell. 

d. What metals could be used for the electrodes in this electrochemical cell? 

e. Suggest two electrolytes for this electrochemical cell. 

f. In which direction will the current flow? 

g. Draw a simple sketch of the complete cell. 

3. For the following cell: 

Sn\Sn'^+\\Ag+\Ag (4.5) 

a. Give the cathode half-reaction. 

b. Give the anode half-reaction. 

c. Give the overall equation for the electrochemical cell. 

d. Draw a simple sketch of the complete cell. 



87 

4.2 The electrolytic celF 
4.2.1 The Electrolytic cell 

In "The Galvanic Cell", we saw that a chemical reaction that involves a transfer of electrons, can be used 
to produce an electric current. In this section, we are going to see whether the 'reverse' process applies. In 
other words, is it possible to use an electric current to force a particular chemical reaction to occur, which 
would otherwise not take place? The answer is 'yes', and the type of cell that is used to do this, is called an 
electrolytic cell. 

Definition 4.7: Electrolytic cell 

An electrolytic cell is a type of cell that uses electricity to drive a non-spontaneous reaction. 

An electrolytic cell is activated by applying an electrical potential across the anode and cathode to force 
an internal chemical reaction between the ions that are in the electrolyte solution. This process is called 
electrolysis. 

Definition 4.8: Electrolysis 

In chemistry and manufacturing, electrolysis is a method of separating bonded elements and 
compounds by passing an electric current through them. 

4.2.1.1 Demonstration : The movement of coloured ions 

A piece of filter paper is soaked in an ammonia-ammonium chloride solution and placed on a microscope 
slide. The filter paper is then connected to a supply of electric current using crocodile clips and connecting 
wire as shown in the diagram below. A line of copper chromate solution is placed in the centre of the filter 
paper. The colour of this solution is initially green-brown. 



Image notjinished 



Figure 4.2 



The current is then switched on and allowed to run for about 20 minutes. After this time, the central 
coloured band disappears and is replaced by two bands, one yellow and the other blue, which seem to have 
separated out from the first band of copper chromate. 

Explanation: 

• The cell that is used to supply an electric current sets up a potential difference across the circuit, so 
that one of the electrodes is positive and the other is negative. 

• The chromate (Cr04~) ions in the copper chromate solution are attracted to the positive electrode, 
while the Cu^"*" ions are attracted to the negative electrode. 

Conclusion: 

The movement of ions occurs because the electric current in the outer circuit sets up a potential difference 
between the two electrodes. 

Similar principles apply in the electrolytic cell, where substances that are made of ions can be broken 
down into simpler substances through electrolysis. 

4.2.1.2 The electrolysis of copper sulphate 

There are a number of examples of electrolysis. The electrolysis of copper sulphate is just one. 



^This content is available online at <http://siyavula.cnx.Org/content/m39419/l.l/>. 



88 CHAPTER 4. ELECTROCHEMICAL REACTIONS 

4.2.1.2.1 Demonstration : The electrolysis of copper sulphate 

Two copper electrodes are placed in a solution of blue copper sulphate and are connected to a source of 
electrical current as shown in the diagram below. The current is turned on and the reaction is left for a 
period of time. 

Image notjinished 

Figure 4.3 



Observations: 

• The initial blue colour of the solution remains unchanged. 

• It appears that copper has been deposited on one of the electrodes but dissolved from the other. 

Explanation: 

• At the negative cathode, positively charged Cu^+ ions are attracted to the negatively charged electrode. 
These ions gain electrons and are reduced to form copper metal, which is deposited on the electrode. 
The half-reaction that takes place is as follows: Cm^"*" {aq) + 2er -^ Cu (s) (reduction half reaction) 

• At the positive anode, copper metal is oxidised to form Cu^+ ions. This is why it appears that 
some of the copper has dissolved from the electrode. The half-reaction that takes place is as follows: 
Cu (s) -^ Cv?'~^ {aq) + 2e~ (oxidation half reaction) 

• The amount of copper that is deposited at one electrode is approximately the same as the amount of 
copper that is dissolved from the other. The number of Cu^+ ions in the solution therefore remains 
almost the same and the blue colour of the solution is unchanged. 

Conclusion: 

In this demonstration, an electric current was used to split CUSO4 into its component ions, Cu^^ and 
S04~. This process is called electrolysis. 

4.2.1.3 The electrolysis of water 

Water can also undergo electrolysis to form hydrogen gas and oxygen gas according to the following reaction: 

2H2O (0 ^ 2H2 (5) + O2 {g) 

This reaction is very important because hydrogen gas has the potential to be used as an energy source. 
The electrolytic cell for this reaction consists of two electrodes (normally platinum metal) , submerged in an 
electrolyte and connected to a source of electric current. 

The reduction half-reaction that takes place at the cathode is as follows: 

2H2O (0 + 2e- -^ H2 (5) + 20H- {aq) 

The oxidation half-reaction that takes place at the anode is as follows: 

2H2O {I) -^ O2 {g) + 4i7+ {aq) + 46" 

4.2.1.4 A comparison of galvanic and electrolytic cells 

It should be much clearer now that there are a number of differences between a galvanic and an electrolytic 
cell. Some of these differences have been summarised in Table 4.1. 



89 



Item 


Galvanic cell 


Electrolytic cell 


Metals used for electrode 


Two metals with different reac- 
tion potentials are used as elec- 
trodes 


The same metal can be used for 
both the cathode and the anode 


Charge of the anode 


negative 


positive 


Charge of the cathode 


positive 


negative 


The electrolyte solution/s 


The electrolyte solutions are kept 
separate from one another, and 
are connected only by a salt 
bridge 


The cathode and anode are in the 
same electrolyte 


Energy changes 


Chemical potential energy from 
chemical reactions is converted to 
electrical energy 


An external supply of electrical 
energy causes a chemical reaction 
to occur 


Applications 


Run batteries, electroplating 


Electrolysis e.g. of water, NaCl 



Table 4.1: A comparison of galvanic and electrolytic cells 



4.2.1.4.1 Electrolyis 



1. An electrolytic cell consists of two electrodes in a silver chloride (AgCl) solution, connected to a source 
of current. A current is passed through the solution and Ag"*" ions are reduced to a silver metal deposit 
on one of the electrodes. 

a. Give the equation for the reduction half-reaction. 

b. Give the equation for the oxidation half-reacion. 

2. Electrolysis takes place in a solution of molten lead bromide (PbBr) to produce lead atoms. 

a. Draw a simple diagram of the electrolytic cell. 

b. Give equations for the half-reactions that take place at the anode and cathode, and include these 
in the diagram. 

c. On your diagram, show the direction in which current flows. 



4.3 Standard potentials^ 

4,3.1 Standard electrode potentials 

The voltages recorded earlier when zinc and copper were connected to a standard hydrogen electrode are in 
fact the standard electrode potentials for these two metals. It is important to remember that these are 
not absolute values, but are potentials that have been measured relative to the potential of hydrogen if the 
standard hydrogen electrode is taken to be zero. 

TIP: By convention, the hydrogen electrode is written on the left hand side of the cell. The sign 
of the voltage tells you the sign of the metal electrode. 

In the examples we used earlier, zinc's electrode potential is actually -0.76 and copper is +0.34. So, if a metal 
has a negative standard electrode potential, it means it forms ions easily. The more negative the value, the 
easier it is for that metal to form ions. If a metal has a positive standard electrode potential, it means it 
does not form ions easily. This will be explained in more detail below. 



^This content is available online at <http://siyavula.cnx.Org/content/m39420/l.l/>. 



90 



CHAPTER 4. ELECTROCHEMICAL REACTIONS 



Luckily for us, we do not have to calculate the standard electrode potential for every metal. This has 
been done already and the results are recorded in a table of standard electrode potentials (Table 4.2). 



Half- Reaction 


E"V 


Li+ + e- ^ Li 


-3.04 


K+ + e" ^K 


-2.92 


Ba^+ + 2e- ^ Ba 


-2.90 


Ca'^+ + 2e- ^ Ca 


-2.87 


Na+ + e^ ^ Na 


-2.71 


Mg'^+ + 2e- ^ Mg 


-2.37 


Mn^+ + 2e" ^ Mn 


-1.18 


2H20 + 2e" ^ H2 [g] + 20H- 


-0.83 


Zn^+ + 2e- ^ Zn 


-0.76 


Cr^+ + 2e- ^ Cr 


-0.74 


Fe^+ + 2e- ^ Fe 


-0.44 


Cr^+ + 3e- ^ Cr 


-0.41 


C<f+ + 2e- ^ Cd 


-0.40 


C6'+ + 2e- ^ Co 


-0.28 


Ni^+ + 2e- ^ Ni 


-0.25 


Sn^+ + 2e- ^ Sn 


-0.14 


Pb^+ + 2e- ^ Pb 


-0.13 


Fe^+ + 3e" ^ Fe 


-0.04 


2H+ + 2e- ^ H2 [g) 


0.00 


S + 2H+ + 2e- ^ H2S {g) 


0.14 


Sn^+ + 2e- ^ Sn^+ 


0.15 


Cu^+ + e- ^ Cu+ 


0.16 


SOI+ + AH+ + 2e^ ^ SO2 (5) + 2H2O 


0.17 


Cu'^+ + 2e- ^ Cu 


0.34 


continued on next page 



91 



2H2O + 02 + 4e- ^ AOH- 


0.40 


Cu+ + e- ^Cu 


0.52 


I2 + 2e- ^ 21- 


0.54 


O2 {9) + 2H+ + 2e- ^ H2O2 


0.68 


Fe^+ + e- ^ Fe'+ 


0.77 


NO3 + 2H+ + e- ^ NO2 {g) + H2O 


0.78 


Hg^+ + 2e- ^ Hg (1) 


0.78 


Ag+ + e- ^Ag 


0.80 


NO^ + AH+ + 3e- ^ NO [g) + 2H2O 


0.96 


Br2 + 2e- ^ 2Br- 


1.06 


O2 (<?) + AH+ + 4e- ^ 2H2O 


1.23 


Mn02 + AH+ + 2e- ^ Mn^+ + 2H2O 


1.28 


Cr20'^- + UH+ + 6e- ^ 2Cr^+ + 7H2O 


1.33 


CI2 + 2e- ^ 2Cl- 


1.36 


Au^+ + 3e" ^ ^M 


1.50 


MnO^ + 8i7+ + 5e- ^ Mn^+ + 4i720 


1.52 


Co3+ + e- ^ Co2+ 


1.82 


F2 + 2e- ^ 2i^- 


2.87 



Table 4.2: Standard Electrode Potentials 

A few examples from the table are shown in Table 4.3. These will be used to explain some of the trends 
in the table of electrode potentials. 



Half-Reaction 


E"V 


Li+ +e- ^ Li 


-3.04 


Zn'^+ + 2e- ^ Zn 


-0.76 


Fe^+ + 3e" ^ Fe 


-0.04 


2H+ + 2e- ^ H2 [g] 


0.00 


Cu^+ + 2e- ^ Cu 


0.34 


Hg^+ + 2e- ^ Hg [l) 


0.78 


Ag+ + e- ^ Ag 


0.80 



Table 4.3: A few examples from the table of standard electrode potentials 
Refer to Table 4.3 and notice the following trends: 

• Metals at the top of series (e.g. Li) have more negative values. This means they ionise easily, in other 
words, they release electrons easily. These metals are easily oxidised and are therefore good reducing 
agents. 

• Metal ions at the bottom of the table are good at picking up electrons. They are easily reduced and 
are therefore good oxidising agents. 



92 CHAPTER 4. ELECTROCHEMICAL REACTIONS 

• The reducing ability (i.e. the ability to act as a reducing agent) of the metals in the table increases as 
you move up in the table. 

• The oxidising ability of metals increases as you move down in the table. 

Exercise 4.2: Using the table of Standard Electrode Potentials (Solution on p. 103.) 

The following half-reactions take place in an electrochemical cell: 
Cu2+ + 2e- ^ Cu 
Ag^ + e~ ^ Ag 

1. Which of these reactions will be the oxidation half- reaction in the cell? 

2. Which of these reactions will be the reduction half-reaction in the cell? 



TIP: Before you tackle this problem, make sure you understand exactly what the question is 
asking. If magnesium is able to displace silver from a solution of silver nitrate, this means that 
magnesium metal will form magnesium ions and the silver ions will become silver metal. In other 
words, there will now be silver metal and a solution of magnesium nitrate. This will only happen 
if magnesium has a greater tendency than silver to form ions. In other words, what the question is 
actually asking is whether magnesium or silver forms ions more easily. 

Exercise 4.3: Using the table of Standard Electrode Potentials (Solution on p. 103.) 

Is magnesium able to displace silver from a solution of silver nitrate? 



4.3.1.1 Table of Standard Electrode Potentials 

1. In your own words, explain what is meant by the 'electrode potential' of a metal. 

2. Give the standard electrode potential for each of the following metals: 

a. magnesium 

b. lead 

c. nickel 

3. Refer to the electrode potentials in Table 4.3. 

a. Which of the metals is most likely to be oxidised? 

b. Which metal is most likely to be reduced? 

c. Which metal is the strongest reducing agent? 

d. In the copper half-reaction, does the equilibrium position for the reaction lie to the left or to the 
right? Explain your answer. 

e. In the mercury half-reaction, does the equilibrium position for the reaction lie to the left or to 
the right? Explain your answer. 

f. If silver was added to a solution of copper sulphate, would it displace the copper from the copper 
sulphate solution? Explain your answer. 

4. Use the table of standard electrode potentials to put the following in order from the strongest oxidising 
agent to the weakest oxidising agent. 

• Cu2+ 

• MnOj 

• Bra 

• Zn2+ 

5. Look at the following half- reactions. 

• Ca^+ + 2e- -^ Ca 



93 

• Ch + 2e- -^ 2Cl 

• Fe^+ + 3e- -^ Fe 

• I2 + 2e- -^ 2I~ 

a. Which substance is the strongest oxidising agent? 

b. Which substance is the strongest reducing agent? 

6. Which one of the substances listed below acts as the oxidising agent in the following reaction? 3SO2 + 
CrzO^- + 2H+ -^ 3S0^- + 2Cr3+ + H2O 

a. H+ 

b. Cr3+ 

c. SO2 

d. CraO?- 

(lEB Paper 2, 2004) 

7. If zinc is added to a solution of magnesium sulphate, will the zinc displace the magnesium from the 
solution? Give a detailed explanation for your answer. 



4.3.2 Combining half cells 

Let's stay with the example of the zinc and copper half cells. If we combine these cells as we did earlier in 
the chapter ("The Galvanic Cell"), the following two equilibria exist: 

Zn^+ + 2e- ^ Zn {E^ = -0.76V") 

Cu2+ + 2e- ^ Cu {E^ = +QMV) 

We know from demonstrations, and also by looking at the sign of the electrode potential, that when 
these two half cells are combined, zinc will be the oxidation half-reaction and copper will be the reduction 
half-reaction. A voltmeter connected to this cell will show that the zinc electrode is more negative than the 
copper electrode. The reading on the meter will show the potential difference between the two half cells. 
This is known as the electromotive force (emf ) of the cell. 

Definition 4.9: Electromotive Force (emf) 

The emf of a cell is defined as the maximum potential difference between two electrodes or half 
cells in a voltaic cell, emf is the electrical driving force of the cell reaction. In other words, the 
higher the emf, the stronger the reaction. 

Definition 4.10: Standard emf (E°g;,) 

Standard emf is the emf of a voltaic cell operating under standard conditions (i.e. 100 kPa, 
concentration = 1 mol.dm""^ and temperature = 298 K). The symbol ° denotes standard conditions. 

When we want to represent this cell, it is shown as follows: 

Zn\Zn'^+ {Imol.dm-^) \\Cu^+ {imol.dm-^) \Cu (4.6) 

The anode half cell (where oxidation takes place) is always written on the left. The cathode half cell 
(where reduction takes place) is always written on the right. 

It is important to note that the potential difference across a cell is related to the extent to which 
the spontaneous cell reaction has reached equilibrium. In other words, as the reaction proceeds and the 
concentration of reactants decreases and the concentration of products increases, the reaction approaches 
equilibrium. When equilibrium is reached, the emf of the cell is zero and the cell is said to be 'fiat'. There 
is no longer a potential difference between the two half cells, and therefore no more current will flow. 

4.3.3 Uses of standard electrode potential 

Standard electrode potentials have a number of different uses. 



94 CHAPTER 4. ELECTROCHEMICAL REACTIONS 

4.3.3.1 Calculating the emf of an electrochemical cell 

To calculate the emf of a cell, you can use any one of the following equations: 

E9 gj„ = E" (right) - E° (left) ('right' refers to the electrode that is written on the right in standard cell 
notation. 'Left' refers to the half-reaction written on the left in this notation) 

E9 gjjN = E'' (reduction half reaction) - E° (oxidation half reaction) 

E9 jjx = E° (oxidising agent) - E" (reducing agent) 

^tceii) = ^° (cathode) - E° (anode) 
So, for the Zn-Cu cell, 

Kelt) = 0-34 - (-0-76) 
= 0.34 + 0.76 
= 1.1 V 

Exercise 4.4: Calculating the emf of a cell (Solution on p. 103.) 

The following reaction takes place: 

Cu (s) + Ag+ {aq) -^ Cu^+ {aq) + Ag (s) 

1. Represent the cell using standard notation. 

2. Calculate the cell potential (emf) of the electrochemical cell. 

Exercise 4.5: Calculating the emf of a cell (Solution on p. 103.) 

Calculate the cell potential of the electrochemical cell in which the following reaction takes place, 
and represent the cell using standard notation. 
Mg (s) + 2H+ {aq) -^ Mg2 + {aq) + H^ {g) 



4.3.3.2 Predicting whether a reaction will take place spontaneously 

Look at the following example to help you to understand how to predict whether a reaction will take place 
spontaneously or not. 

In the reaction, 

P62+ {aq) + 2Br- {aq) -^ Br2 {I) + Pb (s) 

the two half reactions are as follows: 

PP+ + 2er ^ Ph (-0.13 V) 

Br^ -h 2e- ^ 2Br- (+1.06 V) 

TIP: Half cell reactions 

You will see that the half reactions are written as they appear in the table of standard electrode potentials. 
It may be useful to highlight the reacting substance in each half reaction. In this case, the reactants are 
Pb^+ and Br~ ions. 

Look at the electrode potential for the first half reaction. The negative value shows that lead loses 
electrons easily, in other words it is easily oxidised. The reaction would normally proceed from right to left 
(i.e. the equilibrium lies to the left), but in the original equation, the opposite is happening. It is the Pb^+ 
ions that are being reduced to lead. This part of the reaction is therefore not spontaneous. The positive 
electrode potential value for the bromine half-reaction shows that bromine is more easily reduced, in other 
words the equilibrium lies to the right. The spontaneous reaction proceeds from left to right. This is not 
what is happening in the original equation and therefore this is also not spontaneous. Overall it is clear then 
that the reaction will not proceed spontaneously. 

Exercise 4.6: Predicting w^hether a reaction is spontaneous (Solution on p. 104.) 

Will copper react with dilute sulfuric acid (H2SO4)? You are given the following half reactions: 
Cu^+ {aq) + 2e- ^ Cu (s) (E" = +0.34 V) 
2H+ {aq) + 26" ^ H2 {g) (E° = V) 



95 



TIP: 

A second method for predicting whether a reaction is spontaneous 

Another way of predicting whether a reaction occurs spontaneously, is to look at the sign of the emf 
value for the cell. If the emf is positive then the reaction is spontaneous. If the emf is negative, then 
the reaction is not spontaneous. 

4.3.3.3 Balancing redox reactions 

We will look at this in more detail in the next section. 

4.3.3.3.1 Predicting whether a reaction w^ill take place spontaneously 

1. Predict whether the following reaction will take place spontaneously or not. Show all your working. 

2Ag (s) + Cu'^+ (aq) -^ Cu (s) + 2Ag+ (aq) (4.7) 

2. Zinc metal reacts with an acid, H+ (aq) to produce hydrogen gas. 

a. Write an equation for the reaction, using the table of electrode potentials. 

b. Predict whether the reaction will take place spontaneously. Show your working. 

3. Four beakers are set up, each of which contains one of the following solutions: 

a. Mg(N03)2 

b. Ba(N03)2 

C. Cu(N03)2 

d. A1(N03)2 

Iron is added to each of the beakers. In which beaker will a spontaneous reaction take place? 

4. Which one of the following solutions can be stored in an aluminium container? 

a. Cu(S0)4 

b. Zn(S0)4 

c. NaCl 

d. Pb(N03)2 

4.3.3.3.2 Electrochemical cells and standard electrode potentials 

1. An electrochemical cell is made up of a copper electrode in contact with a copper nitrate solution 
and an electrode made of an unknown metal M in contact with a solution of MNO3. A salt bridge 
containing a KNO3 solution joins the two half cells. A voltmeter is connected across the electrodes. 
Under standard conditions the reading on the voltmeter is 0.46V. 



Image notjinished 



Figure 4.4 



The reaction in the copper half cell is given by: Cu -^ Cu^^ + 2e 

a. Write down the standard conditions which apply to this electrochemical cell. 



96 CHAPTER 4. ELECTROCHEMICAL REACTIONS 

b. Identify the metal M. Show calculations. 

c. Use the standard electrode potentials to write down equations for the: 

1. cathode half- reaction 

2. anode half- reaction 

3. overall cell reaction 

d. What is the purpose of the salt bridge? 

e. Explain why a KCl solution would not be suitable for use in the salt bridge in this cell. 

(lEB Paper 2, 2004) 

2. Calculate the emf for each of the following standard electrochemical cells: 

a. 

Mg\Mg^+\\H+\H2 (4.8) 

b. 

Fe\Fe^+\\Fe^+\Fe (4.9) 

c. 

Cr\Cr2 + \\Cu'^+\Cu (4.10) 

d. 

Pb\Pb'^+\\Hg'^+\Hg (4.11) 

3. Given the following two half-reactions: 

• Fe^+ {aq) + e' ^ Fe'^+ [aq) 



• 



MnO^ (aq) + 8H+ (aq) + Se" ^ Mn^+ {aq) + AH2O [l) 



a. Give the standard electrode potential for each half-reaction. 

b. Which reaction takes place at the cathode and which reaction takes place at the anode? 

c. Represent the electrochemical cell using standard notation. 

d. Calculate the emf of the cell 



4.4 Balancing redox reactions* 

4.4.1 Balancing redox reactions 

Half reactions can be used to balance redox reactions. We are going to use some worked examples to help 
explain the method. 

Exercise 4.7: Balancing redox reactions (Solution on p. 104.) 

Magnesium reduces copper (II) oxide to copper. In the process, magnesium is oxidised to magne- 
sium ions. Write a balanced equation for this reaction. 

Exercise 4.8: Balancing redox reactions (Solution on p. 104.) 

Chlorine gas oxidises Fe(II) ions to Fe(III) ions. In the process, chlorine is reduced to chloride 
ions. Write a balanced equation for this reaction. 

Exercise 4.9: Balancing redox reactions in an acid medium (Solution on p. 105.) 

The following reaction takes place in an acid medium: 
Cr20^~ + H2S^ Cr^+ + S 
Write a balanced equation for this reaction. 



*This content is available online at <http://siyavula.cnx.Org/content/m39422/l.l/>. 



97 



Exercise 4.10: Balancing redox reactions in an alkaline medium (Solution on p. 105.) 

K ammonia solution is added to a solution that contains cobalt (II) ions, a complex ion is formed, 
called the hexaaminecobalt(II) ion (Co(NH3)g"'"). In a chemical reaction with hydrogen peroxide so- 
lution, hexaaminecobalt ions are oxidised by hydrogen peroxide solution to the hexaaminecobalt(III) 
ion Co(NH3)g'''. Write a balanced equation for this reaction. 



4.4.1.1 Balancing redox reactions 

1. Balance the following equations. 

a. HNO3 + PbS -^ PhSOA + NO + H2O 

b. Nal + Fe2{SOi)^ ^ h + FeSOi + Na2S0i 

2. Manganate(VII) ions (Mn04 ) oxidise hydrogen peroxide (H2O2) to oxygen gas. The reaction is done 
in an acid medium. During the reaction, the manganate(VII) ions are reduced to manganese(II) ions 
(Mn^+). Write a balanced equation for the reaction. 

3. Chlorine gas is prepared in the laboratory by adding concentrated hydrochloric acid to manganese 
dioxide powder. The mixture is carefully heated. 

a. Write down a balanced equation for the reaction which takes place. 

b. Using standard electrode potentials, show by calculations why this mixture needs to be heated. 

c. Besides chlorine gas which is formed during the reaction, hydrogen chloride gas is given off when 
the conentrated hydrochloric acid is heated. Explain why the hydrogen chloride gas is removed 
from the gas mixture when the gas is bubbled through water. (lEB Paper 2, 2004) 

4. The following equation can be deduced from the table of standard electrode potentials: 2Ct20j^ (aq)-l- 
16H+ (aq) -^ 4Cr^+ (aq) + 3O2 (ff) + SH2O (l) (E° = +0.10V) This equation implies that an acidified 
solution of aqueous potassium dichromate (orange) should react to form Cr^+ (green). Yet aqueous 
laboratory solutions of potassium dichromate remain orange for years. Which ONE of the following 
best explains this? 

a. Laboratory solutions of aqueous potassium dichromate are not acidified 

b. The E° value for this reaction is only +0.10V 

c. The activation energy is too low 

d. The reaction is non-spontaneous 

(lEB Paper 2, 2002) 

5. Sulfur dioxide gas can be prepared in the laboratory by heating a mixture of copper turnings and 
concentrated sulfuric acid in a suitable flask. 

a. Derive a balanced ionic equation for this reaction using the half-reactions that take place. 

b. Give the E° value for the overall reaction. 

c. Explain why it is necessary to heat the reaction mixture. 

d. The sulfur dioxide gas is now bubbled through an aqueous solution of potassium dichromate. 
Describe and explain what changes occur during this process. 

(lEB Paper 2, 2002) 



4.5 Applications' 

4,5.1 Applications of electrochemistry 

Electrochemistry has a number of different uses, particularly in industry. We are going to look at a few 
examples. 



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98 CHAPTER 4. ELECTROCHEMICAL REACTIONS 

4.5.1.1 Electroplating 

Electroplating is the process of using electrical current to coat an electrically conductive object with a thin 
layer of metal. Mostly, this application is used to deposit a layer of metal that has some desired property (e.g. 
abrasion and wear resistance, corrosion protection, improvement of aesthetic qualities etc.) onto a surface 
that doesn't have that property. Electro-refining (also sometimes called electrowinning is electroplating on 
a large scale. Electrochemical reactions are used to deposit pure metals from their ores. One example is the 
elect rorefining of copper. 

Copper plays a major role in the electrical reticulation industry as it is very conductive and is used in 
electric cables. One of the problems though is that copper must be pure if it is to be an effective current 
carrier. One of the methods used to purify copper, is electrowinning. The copper electrowinning process is 
as follows: 

1. Bars of crude (impure) copper containing other metallic impurities is placed on the anodes. 

2. The cathodes are made up of pure copper with few impurities. 

3. The electrolyte is a solution of aqueous CuSOi and H2SO4. 

4. When current passes through the cell, electrolysis takes place. The impure copper anode dissolves to 
form Cu^+ ions in solution. These positive ions are attracted to the negative cathode, where reduction 
takes place to produce pure copper metal. The reactions that take place are as follows: At the anode: 

Cu{s)^Cu^+{aq)+2e- (4.12) 

At the cathode: 

Cu+'^ {aq) + 2e- -^ Cu (s) (> 99%purity) (4.13) 

5. The other metal impurities (Zn, Au, Ag, Fe and Pb) do not dissolve and form a solid sludge at the 
bottom of the tank or remain in solution in the electrolyte. 



Image notjinished 

Figure 4.5: A simplified diagram to illustrate what happens during the electrowinning of copper 

4.5.1.2 The production of chlorine 

Electrolysis can also be used to produce chlorine gas from brine/seawater (NaCl) . This is sometimes referred 
to as the 'Chlor-alkali' process. The reactions that take place are as follows: 
At the anode the reaction is: 

2Cr ^Cl2{g) + 2e- (4.14) 

whereas at the cathode, the following happens: 

2Na+ + 2H2O + 2e~ -^ 2Na+ + 20H- + H2 (4.15) 

The overall reaction is: 

2Na+ + 2H2O + 2Cl- -^ 2Na+ + 20H- + H2 + C'h (4-16) 



99 



Image notjinished 

Figure 4.6: The electrolysis of sodium chloride 



Chlorine is a very important chemical. It is used as a bleaching agent, a disinfectant, in solvents, 
pharmaceuticals, dyes and even plastics such as polyvinlychloride (PVC). 

4.5.1.3 Extraction of aluminium 

Aluminum metal is a commonly used metal in industry where its properties of being both light and strong 
can be utilized. It is also used in the manufacture of products such as aeroplanes and motor cars. The metal 
is present in deposits of bauxite which is a mixture of silicas, iron oxides and hydrated alumina {AI2O3 x 
H2O). 

Electrolysis can be used to extract aluminum from bauxite. The process described below produces 99% 
pure aluminum: 

1. Aluminum is melted along with cryolite (Na^AlFg) which acts as the electrolyte. Cryolite helps to 
lower the melting point and dissolve the ore. 

2. The anode carbon rods provide sites for the oxidation of O^^ and F^ ions. Oxygen and fiourine gas 
are given off at the anodes and also lead to anode consumption. 

3. At the cathode cell lining, the Al^~^ ions are reduced and metal aluminum deposits on the lining. 

4. The AIFq~ electrolyte is stable and remains in its molten state. 

The basic electrolytic reactions involved are as follows: At the cathode: 

Al+^ + 3e- -^ Al{s) (99%purity) (4.17) 



At the anode: 



The overall reaction is as follows: 



202" ^ 02{g) + ie'- (4.18) 



2AI2O3 -^ 4AI + 3O2 (4.19) 

The only problem with this process is that the reaction is endothermic and large amounts of electricity are 
needed to drive the reaction. The process is therefore very expensive. 

This media object is a Flash object. Please view or download it at 

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Figure 4.7 



100 CHAPTER 4. ELECTROCHEMICAL REACTIONS 

4,5.2 Summary 

• An electrochemical reaction is one where either a chemical reaction produces an external voltage, 
or where an external voltage causes a chemical reaction to take place. 

• In a galvanic cell a chemical reaction produces a current in the external circuit. An example is the 
zinc-copper cell. 

• A galvanic cell has a number of components. It consists of two electrodes, each of which is placed 
in a separate beaker in an electrolyte solution. The two electrolytes are connected by a salt bridge. 
The electrodes are connected two each other by an external circuit wire. 

• One of the electrodes is the anode, where oxidation takes place. The cathode is the electrode where 
reduction takes place. 

• In a galvanic cell, the build up of electrons at the anode sets up a potential difference between the two 
electrodes, and this causes a current to flow in the external circuit. 

• A galvanic cell is therefore an electrochemical cell that uses a chemical reaction between two dissimilar 
electrodes dipped in an electrolyte to generate an electric current. 

• The standard notation for a galvanic cell such as the zinc-copper cell is as follows: 

Zn\Zn^+\\Cu^+\Cu (4.20) 

where 

I = a phase boundary (solid/aqueous) 

II = the salt bridge 

• The galvanic cell is used in batteries and in electroplating. 

• An electrolytic cell is an electrochemical cell that uses electricity to drive a non-spontaneous reaction. 
In an electrolytic cell, electrolysis occurs, which is a process of separating elements and compounds 
using an electric current. 

• One example of an electrolytic cell is the electrolysis of copper sulphate to produce copper and sulphate 
ions. 

• Different metals have different reaction potentials. The reaction potential of metals (in other words, 
their ability to ionise), is recorded in a standard table of electrode potential. The more negative 
the value, the greater the tendency of the metal to be oxidised. The more positive the value, the 
greater the tendency of the metal to be reduced. 

• The values on the standard table of electrode potentials are measured relative to the standard hy- 
drogen electrode. 

• The emf of a cell can be calculated using one of the following equations: E9 ^^n = E" (right) - E° (left) 
E9 gjjs = E'^ (reduction half reaction) - E" (oxidation half reaction) E9 ^jn = E° (oxidising agent) - E" 
(reducing agent) E9 ^^x = E*^ (cathode) - E° (anode) 

• It is possible to predict whether a reaction is spontaneous or not, either by looking at the sign of the 
cell's emf or by comparing the electrode potentials of the two half cells. 

• It is possible to balance redox equations using the half-reactions that take place. 

• There are a number of important applications of electrochemistry. These include electroplating, 
the production of chlorine and the extraction of aluminium. 



4.5.2.1 Summary exercise 

1. For each of the following, say whether the statement is true or false. If it is false, re- write the 
statement correctly. 

a. The anode in an electrolytic cell has a negative charge. 

b. The reaction 2KCIO3 -^ 2KC1 -I- 3(92 is an example of a redox reaction. 

c. Lead is a stronger oxidising agent than nickel. 



101 

2. For each of the following questions, choose the one correct answer. 

a. Which one of the following reactions is a redox reaction? 

1. HCl + NaOH -^ NaCl + H2O 

2. AgNOs + Nal -^ Agl + NaNOa 

3. 2FeCl3 + 2H2O + SO2 -^ H2SO4, + 2HCI + 2FeCl2 

4. BaCh + MgSOi -^ MgCh + BaSOi 
(lEB Paper 2, 2003) 

b. Consider the reaction represented by the following equation: Br2(i) + 2/~ -^ '^Br'^ + l2{s) Which 
one of the following statements about this reaction is correct? 

1. bromine is oxidised 

2. bromine acts as a reducing agent 

3. the iodide ions are oxidised 

4. iodine acts as a reducing agent 
(lEB Paper 2, 2002) 

c. The following equations represent two hypothetical half-reactions: X2 + 2e~ ^^ 2X^ (+1.09 V) 
and y+ + e~ ^^ y (-2.80 V) Which one of the following substances from these half-reactions has 
the greatest tendency to donate electrons? 

1. X- 

2. X2 

3. Y 

4. Y+ 

d. Which one of the following redox reactions will not occur spontaneously at room temperature? 

1. Mn + Cu'^+ -^ Mn^+ + Cu 

2. Zn + sol' + AH+ -^ Zn^+ + SO2 + 2H2O 

3. Fe^+ + 3NO2 + SH2O ^ Fe + 3NO^ + 6H+ 

4. 5H2S + 2MnOl + 6H+ ^ 5S + 2Mn^+ + 8H2O 

e. Which statement is CORRECT for a Zn-Cu galvanic cell that operates under standard conditions? 

1. The concentration of the Zn^"*" ions in the zinc half-cell gradually decreases. 

2. The concentration of the Cu^+ ions in the copper half-cell gradually increases. 

3. Negative ions migrate from the zinc half-cell to the copper half-cell. 

4. The intensity of the colour of the electrolyte in the copper half-cell gradually decreases. 
(DoE Exemplar Paper 2, 2008) 

3. In order to investigate the rate at which a reaction proceeds, a learner places a beaker containing 
concentrated nitric acid on a sensitive balance. A few pieces of copper metal are dropped into the 
nitric acid. 

a. Use the relevant half-reactions from the table of Standard Reduction Potentials to derive the 
balanced nett ionic equation for the reaction that takes place in the beaker. 

b. What chemical property of nitric acid is illustrated by this reaction? 

c. List three observations that this learner would make during the investigation. 

(lEB Paper 2, 2005) 

4. The following reaction takes place in an electrochemical cell: 

Cu (s) + 2AgN03 {aq) -^ CuiNO^i)^ [aq) + 2Ag [s] (4.22) 



a. Give an equation for the oxidation half reaction. 

b. Which metal is used as the anode? 

c. Determine the emf of the cell under standard conditions. 



102 CHAPTER 4. ELECTROCHEMICAL REACTIONS 

(lEB Paper 2, 2003) 

5. The nickel-cadmium (NiCad) battery is small and light and is made in a sealed unit. It is used in 
portable appliances such as calculators and electric razors. The following two half reactions occur when 
electrical energy is produced by the cell. Half reaction 1: Cd(s) + 20H~ (aq) -^ Cd(0H)2 (s) + 2e~ 
Half reaction 2: NiO (OH) (s) + H2O {I) + e~ ^ Ni(0H)2 (s) + OH" (aq) 

a. Which half reaction (1 or 2) occurs at the anode? Give a reason for your answer. 

b. Which substance is oxidised? 

c. Derive a balanced ionic equation for the overall cell reaction for the discharging process. 

d. Use your result above to state in which direction the cell reaction will proceed (forward or reverse) 
when the cell is being charged. 

(lEB Paper 2, 2001) 

6. An electrochemical cell is constructed by placing a lead rod in a porous pot containing a solution of 
lead nitrate (see sketch). The porous pot is then placed in a large aluminium container filled with 
a solution of aluminium sulphate. The lead rod is then connected to the aluminium container by a 
copper wire and voltmeter as shown. 



Image notjinished 



Figure 4.8 



a. Define the term reduction. 

b. In which direction do electrons flow in the copper wire? (Al to Pb or Pb to Al) 

c. Write balanced equations for the reactions that take place at... 

1. the anode 

2. the cathode 

d. Write a balanced nett ionic equation for the reaction which takes place in this cell. 

e. What are the two functions of the porous pot? 

f. Calculate the emf of this cell under standard conditions. 



(lEB Paper 2, 2005) 



103 

Solutions to Exercises in Chapter 4 

Solution to Exercise 4.1 (p. 85) 

Step 1. In the standard notation format, the oxidation reaction is written on the left and the reduction reaction 

on the right. So, in this cell, zinc is oxidised and silver ions are reduced. 
Step 2. Oxidation half-reaction: 

Zn -^ Zn'^+ + 2e- 

Reduction half-reaction: 

Ag+ + e- ^ Ag 
Step 3. When you combine the two half-reactions, all the reactants must go on the left side of the equation 

and the products must go on the right side of the equation. The overall equation therefore becomes: 

Zn + Ag+ + e~ ^ Zn^+ + 2e" + Ag 

Note that this equation is not balanced. This will be discussed later in the chapter. 
Step 4. A build up of electrons occurs where oxidation takes place. This is at the zinc electrode. Current will 

therefore flow from the zinc electrode to the silver electrode. 

Solution to Exercise 4.2 (p. 92) 

Step 1. From the table of standard electrode potentials, the electrode potential for the copper half-reaction is 

+0.34 V. The electrode potential for the silver half-reaction is +0.80 V. 
Step 2. Both values are positive, but silver has a higher positive electrode potential than copper. This means 

that silver does not form ions easily, in other words, silver is more likely to be reduced. Copper is more 

likely to be oxidised and to form ions more easily than silver. Copper is the oxidation half-reaction 

and silver is the reduction half-reaction. 

Solution to Exercise 4.3 (p. 92) 

Step 1. The half-reactions are as follows: 
Mg2+ + 2e- ^ Mg 
Ag+ + e- ^Ag 

Step 2. Looking at the electrode potentials for the magnesium and silver reactions: 
For the magnesium half-reaction: E°V = -2.37 
For the silver half-reaction: E°V = 0.80 

This means that magnesium is more easily oxidised than silver and the equilibrium in this half-reaction 
lies to the left. The oxidation reaction will occur spontaneously in magnesium. Silver is more easily 
reduced and the equilibrium lies to the right in this half-reaction. It can be concluded that magnesium 
will displace silver from a silver nitrate solution so that there is silver metal and magnesium ions in 
the solution. 

Solution to Exercise 4.4 (p. 94) 

Step 1. Cu^+ + 2e" ^ Cu (E°V = 0.16V) 

Ag+ + e~ ^Ag (E°V = 0.80V) 
Step 2. Both half-reactions have positive electrode potentials, but the silver half-reaction has a higher positive 

value. In other words, silver does not form ions easily, and this must be the reduction half-reaction. 

Copper is the oxidation half-reaction. Copper is oxidised, therefore this is the anode reaction. Silver 

is reduced and so this is the cathode reaction. 
Step 3. 

Cu\Cu^+ {imol.dm-^) \\Ag+ (imoLdm^^) \Ag (4.23) 

Step 4. E|'^gj,) = E° (cathode) - E° (anode) 
= +0.80 - (+0.34) 
= +0.46 V 

Solution to Exercise 4.5 (p. 94) 



104 CHAPTER 4. ELECTROCHEMICAL REACTIONS 

Step 1. Mg^+ + 2e- ^ Mg (E°V = -2.37) 

2H+ + 2e" ^ H2 (E°V = 0.00) 
Step 2. From the overall equation, it is clear that magnesium is oxidised and hydrogen ions are reduced in this 

reaction. Magnesium is therefore the anode reaction and hydrogen is the cathode reaction. 
Step 3. 

Mg\Mg^+\\H+\H2 (4.24) 

Step 4. E^'^gj,^ = E° (cathode) - E° (anode) 
= 0.00 - (-2.37) 
= +2.37 V 

Solution to Exercise 4.6 (p. 94) 

Step 1. In the first half reaction, the positive electrode potential means that copper does not lose electrons 
easily, in other words it is more easily reduced and the equilibrium position lies to the right. Another 
way of saying this is that the spontaneous reaction is the one that proceeds from left to right, when 
copper ions are reduced to copper metal. 
In the second half reaction, the spontaneous reaction is from right to left. 

Step 2. What you should notice is that in the original reaction, the reactants are copper (Cu) and sulfuric acid 
(2H+). During the reaction, the copper is oxidised and the hydrogen ions are reduced. But from an 
earlier step, we know that neither of these half reactions will proceed spontaneously in the direction 
indicated by the original reaction. The reaction is therefore not spontaneous. 

Solution to Exercise 4.7 (p. 96) 

Step 1. Mg^ Mg^+ 

Step 2. You are allowed to add hydrogen ions (H+) and water molecules if the reaction takes place in an acid 

medium. If the reaction takes place in a basic medium, you can add either hydroxide ions (0H~) or 

water molecules. In this case, there is one magnesium atom on the left and one on the right, so no 

additional atoms need to be added. 
Step 3. Charges can be balanced by adding electrons to either side. The charge on the left of the equation is 

0, but the charge on the right is +2. Therefore, two electrons must be added to the right hand side so 

that the charges balance. The half reaction is now: 

Mg -^ Mg'^+ + le' 
Step 4. The reduction half reaction is: 

Cu2+ ^ Cu 

The atoms balance but the charges don't. Two electrons must be added to the right hand side. 

Cu2+ + 2e- ^ Cu 
Step 5. No multiplication is needed because there are two electrons on either side. 
Step 6. Mg -I- Cu^^ -I- 2e^ -^ Mg^^ -I- Cu -I- 2e^ (The electrons on either side cancel and you get...) 

Mg + Cu^+ -^ Mg^+ + Cu 
Step 7. In this case, it is. 

Solution to Exercise 4.8 (p. 96) 

Step 1. Fe^+ -^ Fe^+ 

Step 2. There is one iron atom on the left and one on the right, so no additional atoms need to be added. 

Step 3. The charge on the left of the equation is +2, but the charge on the right is +3. Therefore, one electron 

must be added to the right hand side so that the charges balance. The half reaction is now: 

Fe^+ -^ Fe^+ + e" 
Step 4. The reduction half reaction is: 

CI2 -^ ci- 

The atoms don't balance, so we need to multiply the right hand side by two to fix this. Two electrons 
must be added to the left hand side to balance the charges. 
CI2 + 2e- -^ 2Cl- 



105 

Step 5. We need to multiply the oxidation half reaction by two so that the number of electrons on either side 
are balanced. This gives: 

2Fe'^+ -^ 2Fe^+ + 2e- 
Step 6. 2Fe^+ + Ch -^ 2Fe^+ + 2Cr 
Step 7. The equation is balanced. 

Solution to Exercise 4.9 (p. 96) 

Step 1. Cr20^~ -^ Cr^+ 

Step 2. We need to multiply the right side by two so that the number of Cr atoms will balance. To balance 

the oxygen atoms, we will need to add water molecules to the right hand side. 

Cr20'^- -^ 2Cr^+ + 7H2O 

Now the oxygen atoms balance but the hydrogens don't. Because the reaction takes place in an acid 

medium, we can add hydrogen ions to the left side. 

Cr20'^- + UH+ -^ 2Cr^+ + 7H2O 
Step 3. The charge on the left of the equation is (-2+14) = +12, but the charge on the right is +6. Therefore, 

six electrons must be added to the left hand side so that the charges balance. The half reaction is now: 

Cr20'^- + UH+ + 6e- -^ 2Cr^+ + 7H2O 
Step 4. The reduction half reaction after the charges have been balanced is: 

S^- ^S + 2e- 
Step 5. We need to multiply the reduction half reaction by three so that the number of electrons on either side 

are balanced. This gives: 

SS-^- ^ 35" + 6e- 
Step 6. Cr20'^~ + UH+ + 33^' ^ 3S + 2Cr^+ + 7H2O 
Step 7. 

Solution to Exercise 4.10 (p. 97) 

Step 1. Co{NH3)l+ -^ Co{NH3)l+ 

Step 2. The number of atoms are the same on both sides. 

Step 3. The charge on the left of the equation is +2, but the charge on the right is +3. One elctron must be 

added to the right hand side to balance the charges in the equation. The half reaction is now: 

Co {NH3f+ ^ Co {NH3)l+ + e- 
Step 4. Although you don't actually know what product is formed when hydrogen peroxide is reduced, the 

most logical product is 0H~. The reduction half reaction is: 

H2O2 -^ OH- 

After the atoms and charges have been balanced, the final equation for the reduction half reaction is: 

H2O2 + 2e- -^ 20H- 
Step 5. We need to multiply the oxidation half reaction by two so that the number of electrons on both sides 

are balanced. This gives: 

2Co {NHs,)l~^ -^ 2Co {NH3)l'^ + 26" 
Step 6. 2Co {NH3)l+ + H2O2 ^ 2Co {NHs)^^ + 20H- 
Step 7. 



106 CHAPTER 4. ELECTROCHEMICAL REACTIONS 



Chapter 5 

The chemical industry 



5.1 Sasor 

5.1.1 Introduction 

The chemical industry has been around for a very long time, but not always in the way we think of it today! 
Dyes, perfumes, medicines and soaps are all examples of products that have been made from chemicals that 
are found in either plants or animals. However, it was not until the time of the Industrial Revolution that 
the chemical industry as we know it today began to develop. At the time of the Industrial Revolution, the 
human population began to grow very quickly and more and more people moved into the cities to live. With 
this came an increase in the need for things like paper, glass, textiles and soaps. On the farms, there was 
a greater demand for fertilisers to help produce enough food to feed all the people in cities and rural areas. 
Chemists and engineers responded to these growing needs by using their technology to produce a variety of 
new chemicals. This was the start of the chemical industry. 

In South Africa, the key event that led to the growth of the chemical industry was the discovery of 
diamonds and gold in the late 1800's. Mines needed explosives so that they could reach the diamonds and 
gold-bearing rock, and many of the main chemical companies in South Africa developed to meet this need 
for explosives. In this chapter, we are going to take a closer look at one of South Africa's major chemical 
companies, Sasol, and will also explore the chloralkali and fertiliser industries. 

5.1.2 Sasol 

Oil and natural gas are important fuel resources. Unfortunately, South Africa has no large oil reserves 
and, until recently, had very little natural gas. One thing South Africa does have however, is large supplies 
of coal. Much of South Africa's chemical industry has developed because of the need to produce oil and gas 
from coal, and this is where Sasol has played a very important role. 

Sasol was established in 1950, with its main aim being to convert low grade coal into petroleum (crude 
oil) products and other chemical feedstocks. A 'feedstock' is something that is used to make another product. 
Sasol began producing oil from coal in 1955. 

NOTE: The first interest in coal chemistry started as early as the 1920's. In the early 1930's a 
research engineer called Etienne Rousseau was employed to see whether oil could be made from coal 
using a new German technology called the Fischer- Tropsch process. After a long time, and after 
many negotiations, Rousseau was given the rights to operate a plant using this new process. As a 
result, the government-sponsored 'South African Coal, Oil and Gas Corporation Ltd' (commonly 
called 'Sasol') was formed in 1950 to begin making oil from coal. A manufacturing plant was 
established in the Free State and the town of Sasolburg developed around this plant. Production 



^This content is available online at <http://siyavula.cnx.Org/content/m39484/l.l/>. 

107 



108 CHAPTER 5. THE CHEMICAL INDUSTRY 

began in 1955. In 1969, the Natref crude oil refinery was established, and by 1980 and 1982 Sasol 
Two and Sasol Three had been built at Secunda. 



5.1.2.1 Sasol today: Technology and production 

Today, Sasol is an oil and gas company with diverse chemical interests. Sasol has three main areas of 
operation: Firstly, coal to liquid fuels technology, secondly the production of crude oil and thirdly the 
conversion of natural gas to liquid fuel. 

1. Coal to liquid fuels Sasol is involved in mining coal and converting it into synthetic fuels, using the 
Fischer- Tropsch technology. Figure 5.1 is a simplified diagram of the process that is involved. 



Image notjinished 

Figure 5.1: The gasification of coal to produce liquid fuels 



Coal gasification is also known as the Sasol/Lurgi gasification process, and involves converting 
low grade coal to a synthesis gas. Low grade coal has a low percentage carbon, and contains other 
impurities. The coal is put under extremely high pressure and temperature in the presence of steam 
and oxygen. The gas that is produced has a high concentration of hydrogen {H2) and carbon monoxide 
(CO). That is why it is called a 'synthesis gas', because it is a mixture of more than one gas. In the 
Sasol Advanced Synthol (SAS) reactors, the gas undergoes a high temperature Fischer- Tropsch 
conversion. Hydrogen and carbon monoxide react under high pressure and temperature and in the 
presence of an iron catalyst, to produce a range of hydrocarbon products. Below is the generalised 
equation for the process. Don't worry too much about the numbers that you see in front of the reactants 
and products. It is enough just to see that the reaction of hydrogen and carbon monoxide (the two 
gases in the synthesis gas) produces a hydrocarbon and water. (2n + 1) 7J2 + nCO -^ C„i72„+2 + nH20 
A range of hydrocarbons are produced, including petrol, diesel, jet fuel, propane, butane, ethylene, 
polypropylene, alcohols and acetic acids. 

TIP: Different types of fuels It is important to understand the difference between types 
of fuels and the terminology that is used for them. The table below summarises some of the 
fuels that will be mentioned in this chapter. 



Compound 


Description 


Petroleum (crude oil) 


A naturally occurring liquid that forms in 
the earth's lithosphere (see Grade 11 notes). 
It is a mixture of hydrocarbons, mostly alka- 
nes, ranging from C5H12 to CisHag. 


continued on next page 



109 



Natural gas 


Natural gas has the same origin as 
petroleum, but is made up of shorter hy- 
drocarbon chains. 


ParafRn wax 


This is made up of longer hydrocarbon 
chains, making it a solid compound. 


Petrol (gasoline) 


A liquid fuel that is derived from petroleum, 
but which contains extra additives to in- 
crease the octane rating of the fuel. Petrol 
is used as a fuel in combustion engines. 


Diesel 


Diesel is also derived from petroleum, but is 
used in diesel engines. 


Liquid Petroleum Gas (LPG) 


LPG is a mixture of hydrocarbon gases, 
and is used as a fuel in heating appHances 
and vehicles. Some LPG mixtures contain 
mostly propane, while others are mostly bu- 
tane. LPG is manufactured when crude oil 
is refined, or is extracted from natural gas 
suppHes in the ground. 


ParafRn 


This is a technical name for the alkanes, 
but refers specifically to the linear alkanes. 
IsoparafEn refers to non-linear (branched) 
alkanes. 


Jet fuel 


A type of aviation fuel designed for use in 
jet engined aircraft. It is an oil-based fuel 
and contains additives such as antioxidants, 
corrosion inhibitors and icing inhibitors. 



Table 5.1 



You will notice in the diagram that Sasol doesn't only produce liquid fuels, but also a variety of 
other chemical products. Sometimes it is the synthetic fuels themselves that are used as feedstocks 
to produce these chemical products. This is done through processes such as hydrocracking and 
steamcracking. Cracking is when heavy hydrocarbons are converted to simpler light hydrocarbons 
(e.g. LPG and petrol) through the breaking of C-C bonds. A heavy hydrocarbon is one that has a 
high number of hydrogen and carbon atoms (more solid), and a light hydrocarbon has fewer hydrogen 
and carbon atoms and is either a liquid or a gas. 

Definition 5.1: Hydrocracking 

Hydrocracking is a cracking process that is assisted by the presence of an elevated partial 
pressure of hydrogen gas. It produces chemical products such as ethane, LPG, isoparafRns, 
jet fuel and diesel. 

Definition 5.2: Steam cracking 

Steam cracking occurs under very high temperatures. During the process, a liquid or gaseous 
hydrocarbon is diluted with steam and then briefly heated in a furnace at a temperature of 
about 850"'''^C. Steam cracking is used to convert ethane to ethylene. Ethylene is a chemical 
that is needed to make plastics. Steam cracking is also used to make propylene, which is an 
important fuel gas. 

2. Production of crude oil Sasol obtains crude oil off the coast of Gabon (a country in West Africa) 
and refines this at the Natref refinery (Figure 5.2). Sasol also sells liquid fuels through a number of 
service stations. 



110 CHAPTER 5. THE CHEMICAL INDUSTRY 

Image notjinished 

Figure 5.2: Crude oil is refined at Sasol's Natref refinery to produce liquid fuels 



3. Liquid fuels from natural gas Sasol produces natural gas in Mozambique and is expanding its 'gas 
to fuel' technology. The gas undergoes a complex process to produce linear-chained hydrocarbons such 
as waxes and paraffins (Figure 5.3). 



Image notjinished 

Figure 5.3: Conversion of natural gas to liquid fuels 



In the autothermal reactor, methane from natural gas reacts with steam and oxygen over an iron- 
based catalyst to produce a synthesis gas. This is a similar process to that involved in coal gasification. 
The oxygen is produced through the fractional distillation of air. 

Definition 5.3: Fractional distillation 

Fractional distillation is the separation of a mixture into its component parts, or fractions. 
Since air is made up of a number of gases (with the major component being nitrogen), frac- 
tional distillation can be used to separate it into these different parts. 

The syngas is then passes through a Sasol Slurry Phase Distillate (SSPD) process. In this process, 
the gas is reacted at far lower temperatures than in the SAS reactors. Apart from hard wax and candle 
wax, high quality diesel can also be produced in this process. Residual gas from the SSPD process 
is sold as pipeline gas while some of the lighter hydrocarbons are treated to produce kerosene and 
paraffin. Ammonia is also produced, which can be used to make fertilisers. 

NOTE: Sasol is a major player in the emerging Southern African natural gas industry, after 
investing 1.2 billion US dollars to develop onshore gas fields in central Mozambique. Sasol has been 
supplying natural gas from Mozambique's Temane field to customers in South Africa since 2004. 



5.1.2.1.1 Sasol processes 

Refer to the diagrams summarising the three main Sasol processes, and use these to answer the following 
questions: 

1. Explain what is meant by each of the following terms: 

a. crude oil 

b. hydrocarbon 

c. coal gasification 

d. synthetic fuel 

e. chemical feedstock 

2. a. What is diesel? 

b. Describe two ways in which diesel can be produced. 

3. Describe one way in which lighter chemical products such as ethylene, can be produced. 

4. Coal and oil play an important role in Sasol's technology. 



Ill 



a. In the table below, summarise the similarities and differences between coal, oil and natural gas 
in terms of how they are formed ('origin'), their general chemical formula and whether they are 
solid, liquid or gas. 





Coal 


Oil 


Natural gas 


Origin 








General chemical formula 








Solid, Hquid or gas 









Table 5.2 

b. In your own words, describe how coal is converted into liquid fuels. 

c. Explain why Sasol's 'coal to liquid fuels' technology is so important in meeting South Africa's fuel 
needs. 

d. Low grade coal is used to produce liquid fuels. What is the main use of higher grade coal in South 
Africa? 



5.1.2.1.2 Case Study : Safety issues and risk assessments 

Safety issues are important to consider when dealing with industrial processes. Read the following extract 
that appeared in the Business report on 6th February 2006, and then discuss the questions that follow. 

Cape Town - Sasol, the petrochemicals group, was Ukely to face prosecution on 10 charges of 
culpable homicide after an explosion at its Secunda plant in 2004 in which 10 people died, a 
Cape Town labour law specialist said on Friday. The specialist, who did not want to be named, 
was speaking after the inquiry into the explosion was concluded last Tuesday. It was convened 
by the labour department. 

The evidence led at the inquiry showed a failure on the part of the company to conduct a proper 
risk assessment and that: Sasol failed to identify hazards associated with a high-pressure gas 
pipeline running through the plant, which had been shut for extensive maintenance work, in the 
presence of hundreds of people and numerous machines, including cranes, fitters, contractors, 
and welding and cutting machines. Because there had never been a risk assessment, the hazard 
of the high-pressure pipeline had never been identified. 

Because Sasol had failed to identify the risk, it did not take any measures to warn people about 
it, mark the line or take precautions. There had also been inadequacy in planning the shutdown 
work. In the face of a barrage of criticism for the series of explosions that year, Sasol embarked 
on a comprehensive programme to improve safety at its operations and appointed Du Pont Safety 
Resources, the US safety consultancy, to benchmark the petrochemical giant's occupational health 
and safety performance against international best practice. 

1. Explain what is meant by a 'risk assessment'. 

2. Imagine that you have been asked to conduct a risk assessment of the Sasol/Lurgi gasification process. 
What information would you need to know in order to do this assessment? 

3. In groups, discuss the importance of each of the following in ensuring the safety of workers in the 
chemical industry: 

• employing experienced Safety, Health and Environment personnel 

• regular training to identify hazards 

• equipment maintenance and routine checks 

4. What other precautions would you add to this list to make sure that working conditions are safe? 



112 



CHAPTER 5. THE CHEMICAL INDUSTRY 



5.1.2.2 Sasol and the environment 

From its humble beginnings in 1950, Sasol has grown to become a major contributor towards the South 
African economy. Today, the industry produces more than 150 000 barrels of fuels and petrochemicals per 
day, and meets more than 40% of South Africa's liquid fuel requirements. In total, more than 200 fuel and 
chemical products are manufactured at Sasolburg and Secunda, and these products are exported to over 70 
countries worldwide. This huge success is largely due to Sasol's ability to diversify its product base. The 
industry has also helped to provide about 170 000 jobs in South Africa, and contributes around R40 billion 
to the country's Gross Domestic Product (GDP). 

However, despite these obvious benefits, there are always environmental costs associated with industry. 
Apart from the vast quantities of resources that are needed in order for the industry to operate, the production 
process itself produces waste products and pollutants. 

5.1.2.2.1 Consumption of resources 

Any industry will always use up huge amounts of resources in order to function effectively, and the chemical 
industry is no exception. In order for an industry to operate, some of the major resources that are needed 
are energy to drive many of the processes, water, either as a coolant or as part of a process and land for 
mining or operations. 

Refer to the data table below which shows Sasol's water use between 2002 and 2005 {Sasol Sustainable 
Development Report 2005), and answer the questions that follow. 



Water use (lOOOm^) 


2002 


2003 


2004 


2005 


River water 


113 722 


124 179 


131 309 


124 301 


Potable water 


15 126 


10 552 


10 176 


10 753 


Total 


157 617 


178 439 


173 319 


163 203 



Table 5.3 

1. Explain what is meant by 'potable' water. 

2. Describe the trend in Sasol's water use that you see in the above statistics. 

3. Suggest possible reasons for this trend. 

4. List some of the environmental impacts of using large amounts of river water for industry. 

5. Suggest ways in which these impacts could be reduced 



5.1.2.2.2 Industry and the environment 

Large amounts of gases and pollutants are released during production, and when the fuels themselves are 
used. Refer to the table below, which shows greenhouse gas and atmospheric pollution data for Sasol between 
2002 and 2005, and then answer the questions that follow. {Source: Sasol Sustainable Development Report 
2005) 



Greenhouse gases and air pollutants (kilotonnes) 


2002 


2003 


2004 


2005 


Carbon dioxide {CO2) 


57 476 


62 873 


66 838 


60 925 


Hydrogen sulfide {H2S) 


118 


105 


102 


89 


Nitrogen oxides {NOx) 


168 


173 


178 


166 


Sulfur dioxide {SO2) 


283 


239 


261 


222 



113 
Table 5.4 

1. Draw line graphs to show how the quantity of each pollutant produced has changed between 2002 and 
2005. 

2. Describe what you see in the graphs, and suggest a reason for this trend. 

3. Explain what is meant by each of the following terms: 

a. greenhouse gas 

b. global warming 

4. Describe some of the possible effects of global warming. 

5. When sulfur dioxide is present in the atmosphere, it may react with water vapour to produce sulfuric 
acid. In the same way, nitrogen dioxide and water vapour react to form nitric acid. These reactions in 
the atmosphere may cause acid rain. Outline some of the possible consequences of acid rain. 

6. Many industries are major contributors towards environmental problems such as global warming, en- 
vironmental pollution, over-use of resources and acid rain. Industries are in a difficult position: On 
one hand they must meet the ever increasing demands of society, and on the other, they must achieve 
this with as little environmental impact as possible. This is a huge challenge. 

• Work in groups of 3-4 to discuss ways in which industries could be encouraged (or in some cases 
forced) to reduce their environmental impact. 

• Elect a spokesperson for each group, who will present your ideas to the class. 

• Are the ideas suggested by each group practical? 

• How easy or difficult do you think it would be to implement these ideas in South Africa? 

NOTE: Sasol is very aware of its responsibility towards creating cleaner fuels. From 1st January 
2006, the South African government enforced a law to prevent lead from being added to petrol. 
Sasol has complied with this. One branch of Sasol, Sasol Technology also has a bio-diesel research 
and development programme focused on developing more environmentally friendly forms of diesel. 
One way to do this is to use renewable resources such as soybeans to make diesel. Sasol is busy 
investigating this new technology. 



5.2 The chloralkali industry' 

5.2.1 The Chloralkali Industry 

The chlorine-alkali (chloralkali) industry is an important part of the chemical industry, and produces chlorine 
and sodium hydroxide through the electrolysis of table salt (NaCl). The main raw material is brine which 
is a saturated solution of sodium chloride (NaCl) that is obtained from natural salt deposits. 

The products of this industry have a number of important uses. Chlorine is used to purify water, and is 
used as a disinfectant. It is also used in the manufacture of many every-day items such as hypochlorous acid, 
which is used to kill bacteria in drinking water. Chlorine is also used in paper production, antiseptics, food, 
insecticides, paints, petroleum products, plastics (such as polyvinyl chloride or PVC), medicines, textiles, 
solvents, and many other consumer products. Many chemical products such as chloroform and carbon 
tetrachloride also contain chlorine. 

Sodium hydroxide (also known as 'caustic soda') has a number of uses, which include making soap and 
other cleaning agents, purifying bauxite (the ore of aluminium), making paper and making rayon (artificial 
silk). 



^This content is available online at <http://siyavula.cnx.Org/content/m39479/l.l/>. 



114 CHAPTER 5. THE CHEMICAL INDUSTRY 

5.2.1.1 The Industrial Production of Chlorine and Sodium Hydroxide 

Chlorine and sodium hydroxide can be produced through a number of different reactions. However, one of 
the problems is that when chlorine and sodium hydroxide are produced together, the chlorine combines with 
the sodium hydroxide to form chlorate {C10~) and chloride {Cl~) ions. This produces sodium chlorate, 
NaClO, a component of household bleach. To overcome this problem the chlorine and sodium hydroxide 
must be separated from each other so that they don't react. There are three industrial processes that have 
been designed to overcome this problem, and to produce chlorine and sodium hydroxide. All three methods 
involve electrolytic cells (chapter ). 

TIP: Electrolytic cells are used to transform reactants into products by using electric current. They 
are made up of an electrolyte and two electrodes, the cathode and the anode. An electrolytic 
cell is activated by applying an external electrical current. This creates an electrical potential across 
the cathode and anode, and forces a chemical reaction to take place in the electrolyte. Cations 
flow towards the cathode and are reduced. Anions flow to the anode and are oxidised. Two new 
products are formed, one product at the cathode and one at the anode. 

1. The Mercury Cell In the mercury-cell (Figure 5.4), brine passes through a chamber which has a 
carbon electrode (the anode) suspended from the top. Mercury flows along the floor of this chamber and 
acts as the cathode. When an electric current is applied to the circuit, chloride ions in the electrolyte 
are oxidised to form chlorine gas. 2C17 s -^ ^2(9) + 2e~ At the cathode, sodium ions are reduced 

to sodium. 2Nat^ ^ + 2e~ -^ 2Na(Hg) The sodium dissolves in the mercury, forming an amalgam of 
sodium and mercury. The amalgam is then poured into a separate vessel, where it decomposes into 
sodium and mercury. The sodium reacts with water in the vessel and produces sodium hydroxide 
(caustic soda) and hydrogen gas, while the mercury returns to the electrolytic cell to be used again. 
2Na(Hg) + 2i/20(,) ^ 2NaOH(,q) + i/2(<,) 



Image notjinished 

Figure 5.4: The Mercury Cell 



This method, however, only produces a fraction of the chlorine and sodium hydroxide that is used by 
industry as it has certain disadvantages: mercury is expensive and toxic, and although it is returned 
to the electrolytic cell, some always escapes with the brine that has been used. The mercury reacts 
with the brine to form mercury(II) chloride. In the past this effluent was released into lakes and rivers, 
causing mercury to accumulate in flsh and other animals feeding on the flsh. Today, the brine is treated 
before it is discharged so that the environmental impact is lower. 
2. The Diaphragm Cell In the diaphragm-cell (Figure 5.5), a porous diaphragm divides the electrolytic 
cell, which contains brine, into an anode compartment and a cathode compartment. The brine is intro- 
duced into the anode compartment and flows through the diaphragm into the cathode compartment. 
When an electric current passes through the brine, the salt's chlorine ions and sodium ions move to 
the electrodes. Chlorine gas is produced at the anode. At the cathode, sodium ions react with water, 
forming caustic soda and hydrogen gas. Some salt remains in the solution with the caustic soda and 
can be removed at a later stage. 



115 

Image notjinished 

Figure 5.5: Diaphragm Cell 



This method uses less energy than the mercury cell, but the sodium hydroxide is not as easily concen- 
trated and precipitated into a useful substance. 

NOTE: To separate the chlorine from the sodium hydroxide, the two half-cells were tradi- 
tionally separated by a porous asbestos diaphragm, which needed to be replaced every two 
months. This was damaging to the environment, as large quantities of asbestos had to be 
disposed. Today, the asbestos is being replaced by other polymers which do not need to be 
replaced as often. 

3. The Membrane Cell The membrane cell (Figure 5.6) is very similar to the diaphragm cell, and 
the same reactions occur. The main difference is that the two electrodes are separated by an ion- 
selective membrane, rather than by a diaphragm. The structure of the membrane is such that it allows 
cations to pass through it between compartments of the cell. It does not allow anions to pass through. 
This has nothing to do with the size of the pores, but rather with the charge on the ions. Brine 
is pumped into the anode compartment, and only the positively charged sodium ions pass into the 
cathode compartment, which contains pure water. 



Image notjinished 

Figure 5.6: Membrane Cell 



At the positively charged anode, CI ions from the brine are oxidised to CI2 gas. 2C1 -^ ^h(g) + 2e 
At the negatively charged cathode, hydrogen ions in the water are reduced to hydrogen gas. 2H'^^ . + 
2e~ -^ H2(g) The Na+ ions flow through the membrane to the cathode compartment and react with 
the remaining hydroxide {OH~) ions from the water to form sodium hydroxide (NaOH). The chloride 
ions cannot pass through, so the chlorine does not come into contact with the sodium hydroxide in 
the cathode compartment. The sodium hydroxide is removed from the cell. The overall equation is as 
follows: 2NaCl -I- 2H2O -^ CI2 -I- H2 + 2NaOH The advantage of using this method is that the sodium 
hydroxide that is produced is very pure because it is kept separate from the sodium chloride solution. 
The caustic soda therefore has very little salt contamination. The process also uses less electricity and 
is cheaper to operate. 

5.2.1.1.1 The Chlor alkali industry 

1. Refer to the flow diagram below which shows the reactions that take place in the membrane cell, and 
then answer the questions that follow. 



Im.age notjinished 

Figure 5.7 



116 



CHAPTER 5. THE CHEMICAL INDUSTRY 



a. What liquid is present in the cathode compartment at (a)? 

b. Identify the gas that is produced at (b). 

c. Explain one feature of this cell that allows the Na"*" and 0H~ ions to react at (c). 

d. Give a balanced equation for the reaction that takes place at (c). 

2. Summarise what you have learnt about the three types of cells in the chloralkali industry by completing 
the table below: 





Mercury cell 


Diaphragm cell 


Membrane cell 


Main raw material 








Mechanism of separating CI2 and NaOH 








Anode reaction 








Cathode reaction 








Purity of NaOH produced 








Energy consumption 








Environmental impact 









Table 5.5 



5.2.1.2 Soaps and Detergents 

Another important part of the chloralkali industry is the production of soaps and detergents. You will 
remember from an earlier chapter, that water has the property of surface tension. This means that it tends 
to bead up on surfaces and this slows down the wetting process and makes cleaning difficult. You can 
observe this property of surface tension when a drop of water falls onto a table surface. The drop holds 
its shape and does not spread. When cleaning, this surface tension must be reduced so that the water can 
spread. Chemicals that are able to do this are called surfactants. Surfactants also loosen, disperse and 
hold particles in suspension, all of which are an important part of the cleaning process. Soap is an example 
of one of these surfactants. Detergents contain one or more surfactants. We will go on to look at these in 
more detail. 

Definition 5.4: Surfactant 

A surfactant is a wetting agent that lowers the surface tension of a liquid, allowing it to spread 
more easily. 

1. Soaps In chapter , a number of important biological macromolecules were discussed, including carbo- 
hydrates, proteins and nucleic acids. Fats are also biological macromolecules. A fat is made up of an 
alcohol called glycerol, attached to three fatty acids (Figure 5.8). Each fatty acid is made up of a 
carboxylic acid attached to a long hydrocarbon chain. An oil has the same structure as a fat, but is a 
liquid rather than a solid. Oils are found in plants (e.g. olive oil, sunflower oil) and fats are found in 
animals. 



Image notjinished 



Figure 5.8: The structure of a fat, composed of an alcohol and three fatty acids 



117 

To make soap, sodium hydroxide (NaOH) or potassium hydroxide (KOH) must be added to a fat or 
an oil. During this reaction, the glycerol is separated from the fatty acid in the fat, and is replaced by 
either potassium or sodium ions (Figure 5.9). Soaps are the water-soluble sodium or potassium salts 
of fatty acids. 

Image notjinished 

Figure 5.9: Sodium hydroxide reacts with a fat to produce glycerol and sodium salts of the fatty acids 



NOTE: Soaps can be made from either fats or oils. Beef fat is a common source of fat, and 
vegetable oils such as palm oil are also commonly used. 

Fatty acids consist of two parts: a carboxylic acid group and a hydrocarbon chain. The hydrocarbon 
chain is hydrophobic, meaning that it is repelled by water. However, it is attracted to grease, oils and 
other dirt. The carboxylic acid is hydrophiUc, meaning that it is attracted to water. Let's imagine 
that we have added soap to water in order to clean a dirty rugby jersey. The hydrocarbon chain will 
attach itself to the soil particles in the jersey, while the carboxylic acid will be attracted to the water. 
In this way, the soil is pulled free of the jersey and is suspended in the water. In a washing machine 
or with vigourous handwashing, this suspension can be rinsed off with clean water. 

Definition 5.5: Soap 

Soap is a surfactant that is used with water for washing and cleaning. Soap is made by 
reacting a fat with either sodium hydroxide (NaOH) or potassium hydroxide (KOH). 

2. Detergents 

Definition 5.6: Detergent 

Detergents are compounds or mixtures of compounds that are used to assist cleaning. The 
term is often used to distinguish between soap and other chemical surfactants for cleaning. 

Detergents are also cleaning products, but are composed of one or more surfactants. Depending on the 
type of cleaning that is needed, detergents may contain one or more of the following: 

• Abrasives to scour a surface. 

• Oxidants for bleaching and disinfection. 

• Enzymes to digest proteins, fats or carbohydrates in stains. These are called biological detergents. 



5.2.1.2.1 The choralkali industry 

Image notjinished 

Figure 5.10 



1. The diagram above shows the sequence of steps that take place in the mercury cell. 

a. Name the 'raw material' in step 1. 

b. Give the chemical equation for the reaction that produces chlorine in step 2. 

c. What other product is formed in step 2. 



118 CHAPTER 5. THE CHEMICAL INDUSTRY 

d. Name the reactants in step 4. 

2. Approximately 30 million tonnes of chlorine are used throughout the world annually. Chlorine is 
produced industrially by the electrolysis of brine. The diagram represents a membrane cell used in the 
production of CI2 gas. 

Image notjinished 

Figure 5.11 



a. What ions are present in the electrolyte in the left hand compartment of the cell? 

b. Give the equation for the reaction that takes place at the anode. 

c. Give the equation for the reaction that takes place at the cathode. 

d. What ion passes through the membrane while these reactions are taking place? Chlorine is used 
to purify drinking water and swimming pool water. The substance responsible for this process is 
the weak acid, hypochlorous acid (HOCl). 

e. One way of putting HOCl into a pool is to bubble chlorine gas through the water. Give an 
equation showing how bubbling Cl2(g) through water produces HOCl. 

f. A common way of treating pool water is by adding 'granular chlorine'. Granular chlorine consists 
of the salt calcium hypochlorite, Ca(0Cl)2. Give an equation showing how this salt dissolves in 
water. Indicate the phase of each substance in the equation. 

g. The 0C1~ ion undergoes hydrolysis , as shown by the following equation: 0C1~ + H2O ^^ 
HOCl + 0H~ Will the addition of granular chlorine to pure water make the water acidic, basic 
or will it remain neutral? Briefly explain your answer. 

(lEB Paper 2, 2003) 



5.3 The fertilizer industry^ 
5.3.1 The Fertiliser Industry 

5.3.1.1 The value of nutrients 

Nutrients are very important for life to exist. An essential nutrient is any chemical element that is needed 
for a plant to be able to grow from a seed and complete its life cycle. The same is true for animals. A 
macronutrient is one that is required in large quantities by the plant or animal, while a micronutrient is one 
that only needs to be present in small amounts for a plant or an animal to function properly. 

Definition 5.7: Nutrient 

A nutrient is a substance that is used in an organism's metabolism or physiology and which must 
be taken in from the environment. 

In plants, the macronutrients include carbon (C), hydrogen (H), oxygen (O), nitrogen (N), phosphorus 
(P) and potassium (K). The source of each of these nutrients for plants, and their function, is summarised 
in Table 5.6. Examples of micronutrients in plants include iron, chlorine, copper and zinc. 



^This content is available online at <http://siyavula.cnx.Org/content/m39482/l.l/>. 



119 



Nutrient 


Source 


Function 


Carbon 


Carbon dioxide in the air 


Component of organic molecules 
such as carbohydrates, lipids and 
proteins 


Hydrogen 


Water from the soil 


Component of organic molecules 


Oxygen 


Water from the soil 


Component of organic molecules 


Nitrogen 


Nitrogen compounds in the soil 


Part of plant proteins and chloro- 
phyll. Also boosts plant growth. 


Phosphorus 


Phosphates in the soil 


Needed for photosynthesis, 
blooming and root growth 


Potassium 


Soil 


Building proteins, part of chloro- 
phyll and reduces diseases in 
plants 



Table 5.6: The source and function of the macronutrients in plants 

Animals need similar nutrients in order to survive. However since animals can't photosynthesise, they 
rely on plants to supply them with the nutrients they need. Think for example of the human diet. We can't 
make our own food and so we either need to eat vegetables, fruits and seeds (all of which are direct plant 
products) or the meat of other animals which would have fed on plants during their life. So most of the 
nutrients that animals need are obtained either directly or indirectly from plants. Table 5.7 summarises the 
functions of some of the macronutrients in animals. 



Nutrient 


Function 


Carbon 


Component of organic compounds 


Hydrogen 


Component of organic compounds 


Oxygen 


Component of organic compounds 


Nitrogen 


Component of nucleic acids and proteins 


Phosphorus 


Component of nucleic acids and phospholipids 


Potassium 


Helps in coordination and regulating the water balance in the body 



Table 5.7: The functions of animal macronutrients 

Micronutrients also play an important function in animals. Iron for example, is found in haemoglobin, 
the blood pigment that is responsible for transporting oxygen to all the cells in the body. 

Nutrients then, are essential for the survival of life. Importantly, obtaining nutrients starts with plants, 
which are able either to photosynthesise or to absorb the required nutrients from the soil. It is important 
therefore that plants are always able to access the nutrients that they need so that they will grow and provide 
food for other forms of life. 



5.3.1.2 The Role of fertilisers 

Plants are only able to absorb soil nutrients in a particular form. Nitrogen for example, is absorbed as 
nitrates, while phosphorus is absorbed as phosphates. The nitrogen cycle (Grade 10) describes the 
process that is involved in converting atmospheric nitrogen into a form that can be used by plants. 

However, all these natural processes of maintaining soil nutrients take a long time. As populations grow 
and the demand for food increases, there is more and more strain on the land to produce food. Often, 



120 



CHAPTER 5. THE CHEMICAL INDUSTRY 



cultivation practices don't give the soil enough time to recover and to replace the nutrients that have been 
lost. Today, fertilisers play a very important role in restoring soil nutrients so that crop yields stay high. 
Some of these fertilisers are organic (e.g. compost, manure and fishmeal), which means that they started 
off as part of something living. Compost for example is often made up of things like vegetable peels and 
other organic remains that have been thrown away. Others are inorganic and can be made industrially. The 
advantage of these commercial fertilisers is that the nutrients are in a form that can be absorbed immediately 
by the plant. 

Definition 5.8: Fertiliser 

A fertiliser is a compound that is given to a plant to promote growth. Fertilisers usually provide 
the three major plant nutrients and most are applied via the soil so that the nutrients are absorbed 
by plants through their roots. 

When you buy fertilisers from the shop, you will see three numbers on the back of the packet e.g. 18- 
24-6. These numbers are called the NPK ratio, and they give the percentage of nitrogen, phosphorus 
and potassium in that fertiliser. Depending on the types of plants you are growing, and the way in which 
you would like them to grow, you may need to use a fertiliser with a slightly different ratio. If you want 
to encourage root growth in your plant for example, you might choose a fertiliser with a greater amount 
of phosphorus. Look at the table below, which gives an idea of the amounts of nitrogen, phosphorus and 
potassium there are in different types of fertilisers. Fertilisers also provide other nutrients such as calcium, 
sulfur and magnesium. 



Description 


Grade (NPK %) 


Ammonium nitrate 


34-0-0 


Urea 


46-0-0 


Bone Meal 


4-21-1 


Seaweed 


1-1-5 


Starter fertiHsers 


18-24-6 


Equal NPK fertiHsers 


12-12-12 


High N, low P and medium K fertilisers 


25-5-15 



Table 5.8: Common grades of some fertiliser materials 

5.3.1.3 The Industrial Production of Fertilisers 

The industrial production of fertilisers may involve several processes. 

1. Nitrogen fertilisers Making nitrogen fertilisers involves producing ammonia, which is then reacted 
with oxygen to produce nitric acid. Nitric acid is used to acidify phosphate rock to produce nitrogen 
fertilisers. The flow diagram below illustrates the processes that are involved. Each of these steps will 
be examined in more detail. 



Image notjinished 



Figure 5.12: Flow diagram showing steps in the production of nitrogen fertilisers 



121 

a. The Haber Process The Haber process involves the reaction of nitrogen and hydrogen to 
produce ammonia. Nitrogen is produced through the fractional distillation of air. Fractional 
distillation is the separation of a mixture (remember that air is a mixture of different gases) 
into its component parts through various methods. Hydrogen can be produced through steam 
reforming. In this process, a hydrocarbon such as methane reacts with water to form carbon 
monoxide and hydrogen according to the following equation: CH4 + H2O -^ CO + 37^2 Nitrogen 
and hydrogen are then used in the Haber process. The equation for the Haber process is: N2 (5) + 
3H2 (g) -^ 2NH3 {g) (The reaction takes place in the presence of an iron (Fe) catalyst under 
conditions of 200 atmospheres (atm) and 450-500 degrees Celsius) 

NOTE: The Haber process developed in the early 20th century, before the start of 
World War 1. Before this, other sources of nitrogen for fertilisers had included saltpeter 
(NaNOa) from Chile and guano. Guano is the droppings of seabirds, bats and seals. 
By the 20th century, a number of methods had been developed to 'fix' atmospheric 
nitrogen. One of these was the Haber process, and it advanced through the work of 
two German men, Fritz Haber and Karl Bosch (The process is sometimes also referred 
to as the 'Haber-Bosch process'). They worked out what the best conditions were in 
order to get a high yield of ammonia, and found these to be high temperature and high 
pressure. They also experimented with different catalysts to see which worked best in 
that reaction. During World War 1, the ammonia that was produced through the Haber 
process was used to make explosives. One of the advantages for Germany was that, 
having perfected the Haber process, they did not need to rely on other countries for the 
chemicals that they needed to make them. 

b. The Ostwald Process The Ostwald process is used to produce nitric acid from ammonia. Nitric 
acid can then be used in reactions that produce fertilisers. Ammonia is converted to nitric acid 
in two stages. First, it is oxidised by heating with oxygen in the presence of a platinum catalyst 
to form nitric oxide and water. This step is strongly exothermic, making it a useful heat source. 
4NH3 {g) + 5O2 {g) -^ 4N0 {g) + 6H2O {g) Stage two, which combines two reaction steps, is 
carried out in the presence of water. Initially nitric oxide is oxidised again to yield nitrogen 
dioxide: 2N0 {g) + O2 {g) -^ 2NO2 (g) This gas is then absorbed by the water to produce nitric 
acid. Nitric oxide is also a product of this reaction. The nitric oxide (NO) is recycled, and the 
acid is concentrated to the required strength. 3NO2 (g) + H2O (/) -^ 2HNO3 (aq) + NO (g) 

c. The Nitrophosphate Process The nitrophosphate process involves acidifying phosphate rock 
with nitric acid to produce a mixture of phosphoric acid and calcium nitrate: Ca3(P04)2 + 
6HNO3 + I2H2O -^ 2H3FOi + 3Ca(N03)2 + I2H2O When calcium nitrate and phosphoric 
acid react with ammonia, a compound fertiliser is produced. Ca(N03)2 + 4i/3P04 + 8NH3 -^ 
CaHP04 + 2NH4NO3 + 8(NH4)2HP04 If potassium chloride or potassium sulphate is added, the 
result will be NPK fertiliser. 

d. Other nitrogen fertilisers 

• Urea ((NH2)2CO) is a nitrogen-containing chemical product which is produced on a large 
scale worldwide. Urea has the highest nitrogen content of all solid nitrogeneous fertilisers 
in common use (46.4%) and is produced by reacting ammonia with carbon dioxide. Two 
reactions are involved in producing urea: 

a. 2NH3 + CO2 -^ H2N - COONH4 

b. 7^2^^ - COONH4 -^ (NH2)2CO + H2O 

• Other common fertilisers are ammonium nitrate and ammonium sulphate. Ammonium nitrate 
is formed by reacting ammonia with nitric acid. NH3 -I- HNO3 -^ NH4NO3 Ammonium 
sulphate is formed by reacting ammonia with sulphuric acid. 2NH3 -I- i?2S04 -^ (NH4)2S04 

2. Phosphate fertilisers The production of phosphate fertilisers also involves a number of processes. 
The first is the production of sulfuric acid through the contact process. Sulfuric acid is then used in 
a reaction that produces phosphoric acid. Phosphoric acid can then be reacted with phosphate rock 
to produce triple superphosphates. 



122 CHAPTER 5. THE CHEMICAL INDUSTRY 

a. The production of sulfuric acid Sulfuric acid is produced from sulfur, oxygen and water through 
the contact process. In the first step, sulfur is burned to produce sulfur dioxide. S (s) + O2 {g) -^ 
SO2 {g) This is then oxidised to sulfur trioxide using oxygen in the presence of a vanadium(V) 
oxide catalyst. 2SO2 + O2 (g) -^ 2SO3 {g) Finally the sulfur trioxide is treated with water to 
produce 98-99% sulfuric acid. SO3 (5) + H2O {I) -^ ^3804 (0 

b. The production of phosphoric acid The next step in the production of phosphate fertiliser is 
the reaction of sulfuric acid with phosphate rock to produce phosphoric acid (H3PO4). In this 
example, the phosphate rock is fiuoropatite (Ca5F(P04)3). Ca5i^(P04)3 + 5i72S04 + 8H2O -^ 
5CaS04 + HF + 3i73P04 

c. The production of phosphates and superphosphates When concentrated phosphoric acid reacts 
with ground phosphate rock, triple superphosphate is produced. 3Ca3(P04)2-CaF2-l-127J3P04 -^ 
9Ca(iJ2P04)2 + 3CaF2 

3. Potassium Potassium is obtained from potash, an impure form of potassium carbonate (K2CO3). 
Other potassium salts (e.g. KCl AND K2O) are also sometimes included in fertilisers. 

5.3.1.4 Fertilisers and the Environment: Eutrophication 

Eutrophication is the enrichment of an ecosystem with chemical nutrients, normally by compounds that 
contain nitrogen or phosphorus. Eutrophication is considered a form of pollution because it promotes plant 
growth, favoring certain species over others. In aquatic environments, the rapid growth of certain types of 
plants can disrupt the normal functioning of an ecosystem, causing a variety of problems. Human society 
is impacted as well because eutrophication can decrease the resource value of rivers, lakes, and estuaries 
making recreational activities less enjoyable. Health- related problems can also occur if eutrophic conditions 
interfere with the treatment of drinking water. 

Definition 5.9: Eutrophication 

Eutrophication refers to an increase in chemical nutrients in an ecosystem. These chemical nutrients 
usually contain nitrogen or phosphorus. 

In some cases, eutrophication can be a natural process that occurs very slowly over time. However, it 
can also be accelerated by certain human activities. Agricultural runoff, when excess fertilisers are washed 
off fields and into water, and sewage are two of the major causes of eutrophication. There are a number of 
impacts of eutrophication. 

• A decrease in biodiversity (the number of plant and animal species in an ecosystem) When a system 
is enriched with nitrogen, plant growth is rapid. When the number of plants increases in an aquatic 
system, they can block light from reaching deeper. Plants also consume oxygen for respiration, and if 
the oxygen content of the water decreases too much, this can cause other organisms such as fish to die. 

• Toxicity Sometimes, the plants that flourish during eutrophication can be toxic and may accumulate 
in the food chain. 

NOTE: South Africa's Department of Water Affairs and Forestry has a 'National Eutrophication 
Monitoring Programme' which was set up to monitor eutrophication in impoundments such as 
dams, where no monitoring was taking place. 

Despite the impacts, there are a number of ways of preventing eutrophication from taking place. Cleanup 
measures can directly remove the excess nutrients such as nitrogen and phosphorus from the water. Creating 
buffer zones near farms, roads and rivers can also help. These act as filters and cause nutrients and 
sediments to be deposited there instead of in the aquatic system. Laws relating to the treatment and 
discharge of sewage can also help to control eutrophication. A final possible intervention is nitrogen testing 
and modeling. By assessing exactly how much fertiliser is needed by crops and other plants, farmers can 
make sure that they only apply just enough fertiliser. This means that there is no excess to run off into 
neighbouring streams during rain. There is also a cost benefit for the farmer. 



123 

5.3.1.4.1 Discussion : Dealing with the consequences of eutrophication 

In many cases, the damage from eutrophication is aheady done. In groups, do the following: 

1. List all the possible consequences of eutrophication that you can think of. 

2. Suggest ways to solve these problems, that arise because of eutrophication. 

5.3.1.4.2 Chemical industry: Fertilisers 

Why we need fertilisers 

There is likely to be a gap between food production and demand in several parts of the world by 
2020. Demand is influenced by population growth and urbanisation, as well as income levels and 
changes in dietary preferences. 

The facts are as follows: 

• There is an increasing world population to feed 

• Most soils in the world used for large-scale, intensive production of crops lack the necessary nutrients 
for the crops 

Conclusion: Fertilisers are needed! 

The flow diagram below shows the main steps in the industrial preparation of two important solid 
fertilisers. 



Image notjinished 

Figure 5.13 



1. Write down the balanced chemical equation for the formation of the brown gas. 

2. Write down the name of process Y. 

3. Write down the chemical formula of liquid E. 

4. Write down the chemical formulae of fertilisers C and D respectively. The following extract comes from 
an article on fertilisers: A world without food for its peopleA world with an environment poisoned 
through the actions of man Are two contributing factors towards a disaster scenario. 

5. Write down THREE ways in which the use of fertilisers poisons the environment. 

5.4 Electrochemistry and batteries* 

5.4,1 Electrochemistry and batteries 

You will remember from chapter that a galvanic cell (also known as a voltaic cell) is a type of electrochemical 
cell where a chemical reaction produces electrical energy. The electromotive force (emf ) of a galvanic cell 
is the difference in voltage between the two half cells that make it up. Galvanic cells have a number 
of applications, but one of the most important is their use in batteries. You will know from your own 
experience that we use batteries in a number of ways, including cars, torches, sound systems and cellphones 
to name just a few. 

*This content is available online at <http://siyavula.cnx.Org/content/m39485/l.l/>. 



124 CHAPTER 5. THE CHEMICAL INDUSTRY 

5.4.1.1 How^ batteries w^ork 

A battery is a device in which chemical energy is directly converted to electrical energy. It consists 
of one or more voltaic cells, each of which is made up of two half cells that are connected in series by a 
conductive electrolyte. The voltaic cells are connected in series in a battery. Each cell has a positive electrode 
(cathode), and a negative electrode (anode). These do not touch each other but are immersed in a solid or 
liquid electrolyte. 

Each half cell has a net electromotive force (emf ) or voltage. The voltage of the battery is the difference 
between the voltages of the half-cells. This potential difference between the two half cells is what causes an 
electric current to flow. 

Batteries are usually divided into two broad classes: 

• Primary batteries irreversibly transform chemical energy to electrical energy. Once the supply of 
reactants has been used up, the battery can't be used any more. 

• Secondary batteries can be recharged, in other words, their chemical reactions can be reversed if 
electrical energy is supplied to the cell. Through this process, the cell returns to its original state. 
Secondary batteries can't be recharged forever because there is a gradual loss of the active materials 
and electrolyte. Internal corrosion can also take place. 



5.4.1.2 Battery capacity and energy 

The capacity of a battery, in other words its ability to produce an electric charge, depends on a number of 
factors. These include: 

• Chemical reactions The chemical reactions that take place in each of a battery's half cells will affect 
the voltage across the cell, and therefore also its capacity. For example, nickel-cadmium (NiCd) cells 
measure about 1.2 V, and alkaline and carbon-zinc cells both measure about 1.5 V. However, in other 
cells such as Lithium cells, the changes in electrochemical potential are much higher because of the 
reactions of lithium compounds, and so lithium cells can produce as much as 3 volts or more. The 
concentration of the chemicals that are involved will also affect a battery's capacity. The higher the 
concentration of the chemicals, the greater the capacity of the battery. 

• Quantity of electrolyte and electrode material in cell The greater the amount of electrolyte in 
the cell, the greater its capacity. In other words, even if the chemistry in two cells is the same, a larger 
cell will have a greater capacity than a small one. Also, the greater the surface area of the electrodes, 
the greater will be the capacity of the cell. 

• Discharge conditions A unit called an Ampere hour (Ah) is used to describe how long a battery 
will last. An ampere hour (more commonly known as an amp hour) is the amount of electric charge 
that is transferred by a current of one ampere for one hour. Battery manufacturers use a standard 
method to rate their batteries. So, for example, a 100 Ah battery will provide a current of 5 A for a 
period of 20 hours at room temperature. The capacity of the battery will depend on the rate at which 
it is discharged or used. If a 100 Ah battery is discharged at 50 A (instead of 5 A), the capacity will 
be lower than expected and the battery will run out before the expected 2 hours. The relationship 
between the current, discharge time and capacity of a battery is expressed by Peukert's law: 

Cp = I'^t (5.1) 

In the equation, 'Cp' represents the battery's capacity (Ah), I is the discharge current (A), k is the 
Peukert constant and t is the time of discharge (hours) . 



125 

5.4.1.3 Lead-acid batteries 

In a lead-acid battery, each cell consists of electrodes of lead (Pb) and lead (IV) oxide (Pb02) in an 
electrolyte of sulfuric acid (H2SO4). When the battery discharges, both electrodes turn into lead (II) sulphate 
(PbS04) and the electrolyte loses sulfuric acid to become mostly water. 

The chemical half reactions that take place at the anode and cathode when the battery is discharging 
are as follows: 

Anode (oxidation): Pb(,) + SO^^^^q) ^ PbS04 (s) + Se" (E° = -0.356 V) 

Cathode (reduction): PbOz (s) + ^Ol^(i,^^ + AH+ + 26" ^ PbS04 (s) + 2i/2O(0 (E° = 1.685 V) 

The overall reaction is as follows: 

PbOa (s) + AH+ (aq) + 2S04~ (aq) + Pb (s) -^ 2PbS04 (s) + 2H2O (/) 

The emf of the cell is calculated as follows: 

EMF = E (cathode)- E (anode) 

EMF = +1.685 V - (-0.356 V) 

EMF = +2.041 V 

Since most batteries consist of six cells, the total voltage of the battery is approximately 12 V. 

One of the important things about a lead-acid battery is that it can be recharged. The recharge reactions 
are the reverse of those when the battery is discharging. 

The lead-acid battery is made up of a number of plates that maximise the surface area on which chemical 
reactions can take place. Each plate is a rectangular grid, with a series of holes in it. The holes are filled 
with a mixture of lead and sulfuric acid. This paste is pressed into the holes and the plates are then stacked 
together, with suitable separators between them. They are then placed in the battery container, after which 
acid is added (Figure 5.14). 



Image notjinished 

Figure 5.14: A lead-acid battery 



Lead-acid batteries have a number of applications. They can supply high surge currents, are relatively 
cheap, have a long shelf life and can be recharged. They are ideal for use in cars, where they provide the high 
current that is needed by the starter motor. They are also used in forklifts and as standby power sources 
in telecommunication facilities, generating stations and computer data centres. One of the disadvantages 
of this type of battery is that the battery's lead must be recycled so that the environment doesn't become 
contaminated. Also, sometimes when the battery is charging, hydrogen gas is generated at the cathode and 
this can cause a small explosion if the gas comes into contact with a spark. 

5.4.1.4 The zinc-carbon dry cell 

A simplified diagram of a zinc-carbon cell is shown in Figure 5.15. 



126 CHAPTER 5. THE CHEMICAL INDUSTRY 



Image notjinished 

Figure 5.15: A zinc-carbon dry cell 



A zinc-carbon cell is made up of an outer zinc container, which acts as the anode. The cathode is the 
central carbon rod, surrounded by a mixture of carbon and manganese (IV) oxide (Mn02)- The electrolyte 
is a paste of ammonium chloride (NH4CI). A fibrous fabric separates the two electrodes, and a brass pin in 
the centre of the cell conducts electricity to the outside circuit. 

The paste of ammonium chloride reacts according to the following half-reaction: 

2NH+ (aq) + 26" ^ 2NH3 (5) + ^2 (s) 

The manganese(IV) oxide in the cell removes the hydrogen produced above, according to the following 
reaction: 

2Mn02 (s) + H2 (<7) ^ MnaOs (s) + H2O {I) 

The combined result of these two reactions can be represented by the following half reaction, which takes 
place at the cathode: 

Cathode: 2NH| (aq) + 2Mn02 (s) + 26" -^ Mn203 (s) + 2NH3 {g) + H2O {I) 

The anode half reaction is as follows: 

Anode: Zn (s) -^ Zn^+ + 2e~ 

The overall equation for the cell is: 

Zn (s) + 2Mn02 (s) + 2NHJ -^ Mn203 (s) + H2O + Zn (NH3)2+ (aq) (E^ = 1.5 V) 

Alkaline batteries are almost the same as zinc-carbon batteries, except that the electrolyte is potassium 
hydroxide (KOH), rather than ammonium chloride. The two half reactions in an alkaline battery are as 
follows: 

Anode: Zn (s) + 20H" (aq) -^ Zn(0H)2 (s) + 2e" 

Cathode: 2Mn02 (s) + H2O (l) + 26" -^ Mn203 (s) + 20Il~ (aq) 

Zinc-carbon and alkaline batteries are cheap primary batteries and are therefore very useful in appliances 
such as remote controls, torches and radios where the power drain is not too high. The disadvantages are 
that these batteries can't be recycled and can leak. They also have a short shelf life. Alkaline batteries last 
longer than zinc-carbon batteries. 

NOTE: The idea behind today's common 'battery' was created by Georges Leclanche in France in 
the 1860's. The anode was a zinc and mercury alloyed rod, the cathode was a porous cup containing 
crushed Mn02. A carbon rod was inserted into this cup. The electrolyte was a liquid solution of 
ammonium chloride, and the cell was therefore called a wet cell. This was replaced by the dry cell 
in the 1880's. In the dry cell, the zinc can which contains the electrolyte, has become the anode, 
and the electrolyte is a paste rather than a liquid. 



5.4.1.5 Environmental considerations 

While batteries are very convenient to use, they can cause a lot of damage to the environment. They use lots 
of valuable resources as well as some potentially hazardous chemicals such as lead, mercury and cadmium. 
Attempts are now being made to recycle the different parts of batteries so that they are not disposed of in 
the environment, where they could get into water supplies, rivers and other ecosystems. 



127 

5.4.1.5.1 Electrochemistry and batteries 

A dry cell, as shown in the diagram below, does not contain a liquid electrolyte. The electrolyte in a typical 
zinc-carbon cell is a moist paste of ammonium chloride and zinc chloride. 

( NOTE TO SELF: Insert diagram ) 

The paste of ammonium chloride reacts according to the following half-reaction: 

2NH+ (aq) + 2e- ^ 2NH3 (5) + H2 (ff) (a) 

Manganese(IV) oxide is included in the cell to remove the hydrogen produced during half- reaction (a), 
according to the following reaction: 

2Mn02 (s) + H2 (<7) ^ MnaOs (s) + H2O {I) (b) 

The combined result of these two half-reactions can be represented by the following half reaction: 

2NH| (aq) + 2Mn02 (s) + 26" ^ MnaOg (s) + 2NH3 {g) + H2O {I) (c) 

1. Explain why it is important that the hydrogen produced in half-reaction (a) is removed by the man- 
ganese(IV) oxide. In a zinc-carbon cell, such as the one above, half-reaction (c) and the half-reaction 
that takes place in the Zn/ZrP'^ half-cell, produce an emf of 1,5 V under standard conditions. 

2. Write down the half-reaction occurring at the anode. 

3. Write down the net ionic equation occurring in the zinc-carbon cell. 

4. Calculate the reduction potential for the cathode half-reaction. 

5. When in use the zinc casing of the dry cell becomes thinner, because it is oxidised. When not in use, 
it still corrodes. Give a reason for the latter observation. 

6. Dry cells are generally discarded when 'fiat'. Why is the carbon rod the most useful part of the cell, 
even when the cell is fiat? 

(DoE Exemplar Paper 2, 2007) 

5.4,2 Summary 

• The growth of South Africa's chemical industry was largely because of the mines, which needed 
explosives for their operations. One of South Africa's major chemical companies is Sasol. Other 
important chemical industries in the country are the chloralkali and fertiliser industries. 

• All countries need energy resources such as oil and natural gas. Since South Africa doesn't have either 
of these resources, Sasol technology has developed to convert coal into liquid fuels. 

• Sasol has three main operation focus areas: Firstly, the conversion of coal to liquid fuel, secondly 
the production and refinement of crude oil which has been imported, and thirdly the production of 
liquid fuels from natural gas. 

• The conversion of coal to liquid fuels involves a Sasol/Lurgi gasification process, followed by the 
conversion of this synthesis gas into a range of hydrocarbons, using the Fischer-Tropsch technology 
in SAS reactors. 

• Heavy hydrocarbons can be converted into light hydrcarbons through a process called cracking. Com- 
mon forms of cracking are hydrocracking and steam cracking. 

• With regard to crude oil, Sasol imports crude oil from Gabon and then refines this at the Natref 
refinery. 

• Gas from Mozambique can be used to produce liquid fuels, through two processes: First, the gas must 
pass through an autothermal reactor to produce a synthesis gas. Secondly, this synthesis gas is passed 
through a Sasol Slurry Phase Distillate process to convert the gas to hydrocarbons. 

• All industries have an impact on the environment through the consumption of natural resources such 
as water, and through the production of pollution gases such as carbon dioxide, hydrogen sulfides, 
nitrogen oxides and others. 

• The chloralkali industry produces chlorine and sodium hydroxide. The main raw material is 
brine (NaCl). 

• In industry, electrolytic cells are used to split the sodium chloride into its component ions to produce 
chlorine and sodium hydroxide. One of the challenges in this process is to keep the products of the 



128 CHAPTER 5. THE CHEMICAL INDUSTRY 

electrolytic reaction (i.e. the chlorine and the sodium hydroxide) separate so that they don't react 
with each other. Specially designed electrolytic cells are needed to do this. 

• There are three types of electrolytic cells that are used in this process: mercury cell, the diaphragm 
cell and the membrane cell. 

• The mercury cell consists of two reaction vessels. The first reaction vessel contains a mercury cathode 
and a carbon anode. An electric current passed through the brine produces Cl~ and Na"*" ions. The 
Cl~ ions are oxidised to form chlorine gas at the anode. Na"*" ions combine with the mercury cathode to 
form a sodium-mercury amalgam. The sodium-mercury amalgam passes into the second reaction vessel 
containing water, where the Na+ ions react with hydroxide ions from the water. Sodium hydroxide is 
the product of this reaction. 

• One of the environmental impacts of using this type of cell, is the use of mercury, which is highly 
toxic. 

• In the diaphragm cell, a porous diaphragm separates the anode and the cathode compartments. 
Chloride ions are oxidised to chlorine gas at the anode, while sodium ions produced at the cathode 
react with water to produce sodium hydroxide. 

• The membrane cell is very similar to the diaphragm cell, except that the anode and cathode com- 
partments are separated by an ion-selective membrane rather than by a diaphragm. Brine is only 
pumped into the anode compartment. Positive sodium ions pass through the membrane into the cath- 
ode compartment, which contains water. As with the other two cells, chlorine gas is produced at the 
anode and sodium hydroxide at the cathode. 

• One use of sodium hydroxide is in the production of soaps and detergents, and so this is another 
important part of the chloralkali industry. 

• To make soap, sodium hydroxide or potassium hydroxide react with a fat or an oil. In the reaction, the 
sodium or potassium ions replace the alcohol in the fat or oil. The product, a sodium or potassium 
salt of a fatty acid, is what soap is made of. 

• The fatty acids in soap have a hydrophilic and a hydrophobic part in each molecule, and this helps 
to loosen dirt and clean items. 

• Detergents are also cleaning products, but are made up of a mixture of compounds. They may also 
have other components added to them to give certain characteristics. Some of these additives may be 
abrasives, oxidants or enzymes. 

• The fertiliser industry is another important chemical industry. 

• All plants need certain macronutrients (e.g. carbon, hydrogen, oxygen, potassium, nitrogen and 
phosphorus) and micronutrients (e.g. iron, chlorine, copper and zinc) in order to survive. Fertilisers 
provide these nutrients. 

• In plants, most nutrients are obtained from the atmosphere or from the soil. 

• Animals also need similar nutrients, but they obtain most of these directly from plants or plant prod- 
ucts. They may also obtain them from other animals, whcih may have fed on plants during their life. 

• The fertiliser industry is very important in ensuring that plants and crops receive the correct nutrients 
in the correct quantities to ensure maximum growth. 

• Nitrogen fertilisers can be produced industrially using a number of chemical processes: The Haber 
process reacts nitrogen and hydrogen to produce ammonia; the Ostwald process reacts oxygen 
and ammonia to produce nitric acid; the nitrophosphate process reacts nitric acid with phosphate 
rock to produce compound fertilisers. 

• Phosphate fertilisers are also produced through a series of reactions. The contact process pro- 
duces sulfuric acid. Sulfuric acid then reacts with phosphate rock to produce phosphoric acid, 
after which phosphoric acid reacts with ground phosphate rock to produce fertilisers such as triple 
superphosphate. 

• Potassium is obtained from potash. 

• Fertilisers can have a damaging effect on the environment when they are present in high quantities 
in ecosystems. They can lead to eutrophication. A number of preventative actions can be taken to 



129 

reduce these impacts. 

Another important part of the chemical industry is the production of batteries. 
A battery is a device that changes chemical energy into electrical energy. 

A battery consists of one or more voltaic cells, each of which is made up of two half cells that are 
connected in series by a conductive electrolyte. Each half cell has a net electromotive force (emf) or 
voltage. The net voltage of the battery is the difference between the voltages of the half-cells. This 
potential difference between the two half cells is what causes an electric current to flow. 
A primary battery cannot be recharged, but a secondary battery can be recharged. 
The capacity of a battery depends on the chemical reactions in the cells, the quantity of elec- 
trolyte and electrode material in the cell, and the discharge conditions of the battery. 
The relationship between the current, discharge time and capacity of a battery is expressed by Peuk- 
ert's law: 

Cp = IH (5.2) 

In the equation, 'Cp' represents the battery's capacity (Ah), I is the discharge current (A), k is the 
Peukert constant and t is the time of discharge (hours) . 

Two common types of batteries are lead-acid batteries and the zinc-carbon dry cell. 
In a lead-acid battery, each cell consists of electrodes of lead (Pb) and lead (IV) oxide (Pb02) in an 
electrolyte of sulfuric acid (H2SO4). When the battery discharges, both electrodes turn into lead (II) 
sulphate (PbS04) and the electrolyte loses sulfuric acid to become mostly water. 
A zinc-carbon cell is made up of an outer zinc container, which acts as the anode. The cathode is 
the central carbon rod, surrounded by a mixture of carbon and manganese (IV) oxide (Mn02). The 
electrolyte is a paste of ammonium chloride (NH4CI). A fibrous fabric separates the two electrodes, 
and a brass pin in the centre of the cell conducts electricity to the outside circuit. 
Despite their many advantages, batteries are made of potentially toxic materials and can be damaging 
to the environment. 



5.4.2.1 Summary Exercise 

1. Give one word or term for each of the following descriptions: 

a. A solid organic compound that can be used to produce liquid fuels. 

b. The process used to convert heavy hydrocarbons into light hydrocarbons. 

c. The process of separating nitrogen from liquid air. 

d. The main raw material in the chloralkali industry. 

e. A compound given to a plant to promote growth. 

f. An electrolyte used in lead-acid batteries. 

2. Indicate whether each of the following statements is true or false. If the statement is false, rewrite the 
statement correctly. 

a. The longer the hydrocarbon chain in an organic compound, the more likely it is to be a solid at 
room temperature. 

b. The main elements used in fertilisers are nitrogen, phosphorus and potassium. 

c. A soap molecule is composed of an alcohol molecule and three fatty acids. 

d. During the industrial preparation of chlorine and sodium hydroxide, chemical energy is converted 
to electrical energy. 

3. For each of the following questions, choose the one correct answer from the list provided. 

a. The sequence of processes that best describes the conversion of coal to liquid fuel is: 

1. coal -^ gas purification -^ SAS reactor -^ liquid hydrocarbon 

2. coal -^ autothermal reactor -^ Sasol slurry phase F-T reactor -^ liquid hydrocarbon 

3. coal -^ coal purification -^ synthesis gas -^ oil 

4. coal -^ coal gasification -^ gas purification -^ SAS reactor -^ liquid hydrocarbons 



130 CHAPTER 5. THE CHEMICAL INDUSTRY 

b. The half-reaction that takes place at the cathode of a mercury cell in the chloralkali industry is: 

1. 2Cr -^ CI2 + 2e- 

2. 2Na+ + 2e- -^ 2Na 

3. 2H+ + 26" -^ H2 

4. NaCl + H2O -^ NaOH + HCl 

c. In a zinc-carbon dry cell... 

1. the electrolyte is manganese (IV) oxide 

2. zinc is oxidised to produce electrons 

3. zinc is reduced to produce electrons 

4. manganese (IV) dioxide acts as a reducing agent 

4. Chloralkali manufacturing process The chloralkali (also called 'chlorine-caustic') industry is one 
of the largest electrochemical technologies in the world. Chlorine is produced using three types of 
electrolytic cells. The simplified diagram below shows a membrane cell. 



Image notjinished 

Figure 5.16 



a. Give two reasons why the membrane cell is the preferred cell for the preparation of chlorine. 

b. Why do you think it is advisable to use inert electrodes in this process? 

c. Write down the equation for the half-reaction taking place at electrode M. 

d. Which gas is chlorine gas? Write down only Gas A or Gas B. 

e. Briefly explain how sodium hydroxide forms in this cell. 

(DoE Exemplar Paper 2,2007) 
5. The production of nitric acid is very important in the manufacture of fertilisers. Look at the diagram 
below, which shows part of the fertiliser production process, and then answer the questions that follow. 



Image notjinished 

Figure 5.17 



a. Name the process at (1). 

b. Name the gas at (2). 

c. Name the process at (3) that produces gas (2). 

d. Name the product at (4). 

e. Name two fertilisers that can be produced from nitric acid. 

6. A lead-acid battery has a number of different components. Match the description in Column A with 
the correct word or phrase in Column B. All the descriptions in Column A relate to lead-acid batter- 
ies. 



131 



Column A 


Column B 


The electrode metal 


Lead sulphate 


Electrolyte 


Mercury 


A product of the overall cell reaction 


Electrolytic 


An oxidising agent in the cathode half-reaction 


Lead 


Type of cells in a lead-acid battery 


Sulfuric acid 




Ammonium chloride 




Lead oxide 




Galvanic 



Table 5.9 



132 CHAPTER 5. THE CHEMICAL INDUSTRY 



Chapter 6 

Motion in two dimensions 



6.1 Vertical projectile motion' 

6.1.1 Introduction 

In Grade 10, we studied motion in one dimension and briefly looked at vertical motion. In this chapter 
we will discuss vertical motion and also look at motion in two dimensions. In Grade 11, we studied the 
conservation of momentum and looked at applications in one dimension. In this chapter we will look at 
momentum in two dimensions. 

6.1.2 Vertical Projectile Motion 

In Grade 10, we studied the motion of objects in free fall and we saw that an object in free fall falls with 
gravitational acceleration g. Now we can consider the motion of objects that are thrown upwards and then 
fall back to the Earth. We call this projectile motion and we will only consider the situation where the 
object is thrown straight upwards and then falls straight downwards - this means that there is no horizontal 
displacement of the object, only a vertical displacement. 

6.1.2.1 Motion in a Gravitational Field 

When an object is in the earth's gravitational field, it always accelerates downwards, towards the centre of 
the earth, with a constant acceleration g, no matter whether the object is moving upwards or downwards. 
This is shown in Figure 6.1. 

TIP: Projectiles moving upwards or downwards in the earth's gravitational field always accelerate 
downwards with a constant acceleration g. Note: acceleration means that the velocity changes; it 
either becomes greater or smaller. 



Image notjinished 

Figure 6.1: Objects moving upwards or downwards, always accelerate downwards. 



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133 



134 CHAPTER 6. MOTION IN TWO DIMENSIONS 

This means that if an object is moving upwards, its velocity decreases until it stops {vf = m-s~^). This is 
the maximum height that the object reaches, because after this, the object starts to fall. 

TIP: Projectiles have zero velocity at their greatest height. 

Consider an object thrown upwards from a vertical height ho- We have seen that the object will travel 
upwards with decreasing velocity until it stops, at which point it starts falling. The time that it takes for 
the object to fall down to height ho is the same as the time taken for the object to reach its maximum height 
from height ho- 



Image notjinished 



Figure 6.2: (a) An object is thrown upwards from height ho. (b) After time tm, the object reaches its 
maximum height, and starts to fall, (c) After a time 2fm the object returns to height ho- 



TIP: Projectiles take the same time to go from the point of launch to the greatest height as the 
time they take to fall back to the point of launch. 



6.1.2.2 Equations of Motion 

The equations of motion that were used in Chapter to describe free fall can be used for projectile motion. 
These equations are the same as those equations that were derived in Chapter , but with acceleration from 
gravity: a = g. We use g = 9,8m ■ s~^ for our calculations. 

Remember that when you use these equations, you are dealing with vectors which have magnitude and 
direction. Therefore, you need to decide which direction will be the positive direction so that your vectors 
have the correct signs. 

Vi = initial velocity (m ■ s^^j at time t = Os 

Vf = final velocity (m ■ s~^j at time t 

Ax = vertical displacement (m) 

, , (6.1) 

t = time (s) 



At = time interval (s) 

g = acceleration due to gravity ( 



m ■ s ^ ) 



Vf = Vi + gt 

Ax = ^-^^t 

Ax = Vit + \gt^ 

v) = vf + 2gAx 



(6.2) 



This simulation allows you to fire various objects from a cannon and see where they land. K you aim the 
cannon straight up (angle of 90) then you will have vertical projectile motion. 



135 



Phet simulation for projectile motion 

This media object is a Flash object. Please view or download it at 
<projectile- motion. swf> 



Figure 6.3 



Exercise 6.1: Projectile motion (Solution on p. 155.) 

A ball is thrown upwards with an initial velocity of 10 m-s~^. 

1. Determine the maximum height reached above the thrower's hand. 

2. Determine the time it takes the ball to reach its maximum height. 

Exercise 6.2: Height of a projectile (Solution on p. 155.) 

A cricketer hits a cricket ball from the ground so that it goes directly upwards. If the ball takes, 
10 s to return to the ground, determine its maximum height. 



6.1.2.2.1 Equations of Motion 

1. A cricketer hits a cricket ball straight up into the air. The cricket ball has an initial velocity of 20 
m-s~^. 

a. What height does the ball reach before it stops to fall back to the ground. 

b. How long has the ball been in the air for? 

2. Zingi throws a tennis ball straight up into the air. It reaches a height of 80 cm. 

a. Determine the initial velocity of the tennis ball. 

b. How long does the ball take to reach its maximum height? 

3. A tourist takes a trip in a hot air balloon. The hot air balloon is ascending (moving up) at a velocity 
of 4 m-s~^. He accidentally drops his camera over the side of the balloon's basket, at a height of 20 m. 
Calculate the velocity with which the camera hits the ground. 



Image notjtnished 



Figure 6.4 



6.1.2.3 Graphs of Vertical Projectile Motion 

Vertical projectile motion is the same as motion at constant acceleration. In Grade 10 you learned about the 
graphs for motion at constant acceleration. The graphs for vertical projectile motion are therefore identical 
to the graphs for motion under constant acceleration. 

When we draw the graphs for vertical projectile motion, we consider two main situations: an object 
moving upwards and an object moving downwards. 

If we take the upwards direction as positive then for an object moving upwards we get the graphs shown 
in Figure 6.5. 



136 CHAPTER 6. MOTION IN TWO DIMENSIONS 



Image notjinished 



Figure 6.5: Graphs for an object thrown upwards with an initial velocity Vi. The object takes tm s to 
reach its maximum height of hm m after which it falls back to the ground, (a) position vs. time graph 
(b) velocity vs. time graph (c) acceleration vs. time graph. 



Exercise 6.3: Drawing Graphs of Projectile Motion (Solution on p. 156.) 

Stanley is standing on the a balcony 20 m above the ground. Stanley tosses up a rubber ball with 
an initial velocity of 4,9 m-s~^. The ball travels upwards and then falls to the ground. Draw graphs 
of position vs. time, velocity vs. time and acceleration vs. time. Choose upwards as the positive 
direction. 

Exercise 6.4: Analysing Graphs of Projectile Motion (Solution on p. 158.) 

The graph below (not drawn to scale) shows the motion of tennis ball that was thrown vertically 
upwards from an open window some distance from the ground. It takes the ball 0,2 s to reach its 
highest point before falling back to the ground. Study the graph given and calculate 

1. how high the window is above the ground. 

2. the time it takes the ball to reach the maximum height. 

3. the initial velocity of the ball. 

4. the maximum height that the ball reaches. 

5. the final velocity of the ball when it reaches the ground. 



Image notjinished 

Figure 6.6 



Exercise 6.5: Describing Projectile Motion (Solution on p. 159.) 

A cricket ball is hit by a cricketer and the following graph of velocity vs. time was drawn: 



Image notjinished 



Figure 6.7 



A cricketer hits a cricket ball from the ground and the following graph of velocity vs. time was 
drawn. Upwards was taken as positive. Study the graph and follow the instructions below: 

1. Describe the motion of the ball according to the graph. 

2. Draw a sketch graph of the corresponding displacement-time graph. Label the axes. 

3. Draw a sketch graph of the corresponding acceleration-time graph. Label the axes. 



137 

6.1.2.3.1 Graphs of Vertical Projectile Motion 

1. Amanda throws a tennisball from a height of 1,5m straight up into the air and then lets it fall to the 
ground. Draw graphs of Ax vs t; v vs t and a vs t for the motion of the ball. The initial velocity of 
the tennisball is 2m ■ s~^. Choose upwards as positive. 

2. A bullet is shot straight upwards from a gun. The following graph is drawn. Downwards was chosen 
as positive 

a. Describe the motion of the bullet. 

b. Draw a displacement - time graph. 

c. Draw an acceleration - time graph. 



Image notjinished 



Figure 6.8 



6.2 Conservation of momentum' 

6.2.1 Conservation of Momentum in Two Dimensions 

We have seen in Grade 11 that the momentum of a system is conserved when there are no external forces 
acting on the system. Conversely, an external force causes a change in momentum Ap, with the impulse 
delivered by the force, F acting for a time At given by: 

Ap = F ■ At (6.3) 

The same principles that were studied in applying the conservation of momentum to problems in one 
dimension, can be applied to solving problems in two dimensions. 

The calculation of momentum is the same in two dimensions as in one dimension. The calculation of 
momentum in two dimensions is broken down into determining the x and y components of momentum and 
applying the conservation of momentum to each set of components. 

Consider two objects moving towards each other as shown in Figure 6.9. We analyse this situation by 
calculating the x and y components of the momentum of each object. 



Image notjinished 

Figure 6.9: Two balls collide at point P. 



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138 

6.2.1.1 Before the collision 

Total momentum: 



CHAPTER 6. MOTION IN TWO DIMENSIONS 



a;-component of momentum: 



2/-component of momentum: 



Pii 

Pi2 



rriiVii 

m2Vi2 



Piix = rriiVii^ = niiV^iCosOi 

Pi2x = rn2Ui2x = m2Vi2COs92 

Piiy = miViiy = miViisinOi 

Pi2y = m2Vi2y = m2Vi2Sin02 



(6.4) 



(6.5) 



(6.6) 



6.2.1.2 After the collision 

Total momentum: 



a;-component of momentum: 



y-component of momentum: 



Pfi = miVfi 

Pf2 = m2Vf2 



Pfix = miVfix = niivficoscpi 

Pf2x = m2Vf2x = m2Vf2COS(f)2 



Pfiy = miVfiy = miVfisincpi 

Pf2y = fn2Vf2y = m2Vf2Sin(f)2 



(6.7) 



(6.J 



(6.9) 



6.2.1.3 Conservation of momentum 

The initial momentum is equal to the final momentum: 

Pi =Pf 

Pi = Pil + Pi2 
Pf = Pfl + Pf2 

This forms the basis of analysing momentum conservation problems in two dimensions. 



(6.10) 
(6.11) 



139 



Phet simulation for Momentum 

This media object is a Flash object. Please view or download it at 
<collision-lab.swf> 



Figure 6.10 



Exercise 6.6: 2D Conservation of Momentum (Solution on p. 160.) 

In a rugby game, Player 1 is running with the ball at 5 m-s~^ straight down the field parallel to 
the edge of the field. Player 2 runs at 6 m-s~^ an angle of 60° to the edge of the field and tackles 
Player 1. In the tackle, Player 2 stops completely while Player 1 bounces off Player 2. Calculate 
the velocity (magnitude and direction) at which Player 1 bounces off Player 2. Both the players 
have a mass of 90 kg. 

Exercise 6.7: 2D Conservation of Momentum: II (Solution on p. 162.) 

In a soccer game. Player 1 is running with the ball at 5 m-s"'^ across the pitch at an angle of 75° 
from the horizontal. Player 2 runs towards Player 1 at 6 m-s~^ an angle of 60° to the horizontal 
and tackles Player 1. In the tackle, the two players bounce off each other. Player 2 moves off with a 
velocity in the opposite x-direction of 0.3 m-s~^ and a velocity in the y-direction of 6 m-s~^. Both 
the players have a mass of 80 kg. What is the final total velocity of Player 1? 



6.3 Types of collisions^ 
6.3.1 Types of Collisions 

Two types of collisions are of interest: 

• elastic collisions 

• inelastic collisions 

In both types of collision, total momentum is always conserved. Kinetic energy is conserved for elastic 
collisions, but not for inelastic collisions. 

6.3.1.1 Elastic Collisions 

Definition 6.1: Elastic Collisions 

An elastic collision is a collision where total momentum and total kinetic energy are both conserved. 

This means that in an elastic collision the total momentum and the total kinetic energy before the 
collision is the same as after the collision. For these kinds of collisions, the kinetic energy is not changed 
into another type of energy. 

6.3.1.1.1 Before the Collision 

Figure 6.11 shows two balls rolling toward each other, about to collide: 



^This content is available online at <http://siyavula.cnx.Org/content/m39539/l.l/>. 



140 CHAPTER 6. MOTION IN TWO DIMENSIONS 



Image notjinished 

Figure 6.11: Two balls before they collide. 



Before the balls collide, the total momentum of the system is equal to all the individual momenta added 
together. Ball 1 has a momentum which we call pn and ball 2 has a momentum which we call pi2, it means 
the total momentum before the collision is: 

Pt=Pii+Pi2 (6-12) 

We calculate the total kinetic energy of the system in the same way. Ball 1 has a kinetic energy which we 
call KEii and the ball 2 has a kinetic energy which we call KEi2, it means that the total kinetic energy 
before the collision is: 

KE^ = KE,i + KE,2 (6.13) 



6.3.1.1.2 After the Collision 

Figure 6.12 shows two balls after they have collided: 



Image notjinished 

Figure 6.12: Two balls after they collide. 



After the balls collide and bounce off each other, they have new momenta and new kinetic energies. Like 
before, the total momentum of the system is equal to all the individual momenta added together. Ball 1 
now has a momentum which we call pfi and ball 2 now has a momentum which we call pf2, it means the 
total momentum after the collision is 

Pf = Pfi+Pf2 (6-14) 

Ball 1 now has a kinetic energy which we call KEfi and ball 2 now has a kinetic energy which we call 
KEf2, it means that the total kinetic energy after the collision is: 

KEf = KEfi + KEf2 (6.15) 

Since this is an elastic collision, the total momentum before the collision equals the total momentum after 
the collision and the total kinetic energy before the collision equals the total kinetic energy after the collision. 



141 

Therefore: 

Initial Final 

Pi = Pf 

P^l+P^2 = Pfl+Pf2 , . 

(6.16) 
and 

KE, = KEf 

KEa + KE,2 = KEf I + KEf^ 

Exercise 6.8: An Elastic Collision (Solution on p. 164.) 

Consider a collision between two pool balls. Ball 1 is at rest and ball 2 is moving towards it with 
a speed of 2 m-s~^. The mass of each ball is 0.3 kg. After the balls collide elastically, ball 2 comes 
to an immediate stop and ball 1 moves off. What is the final velocity of ball 1? 

Exercise 6.9: Another Elastic Collision (Solution on p. 164.) 

Consider two 2 marbles. Marble 1 has mass 50 g and marble 2 has mass 100 g. Edward rolls 
marble 2 along the ground towards marble 1 in the positive x-direction. Marble 1 is initially at rest 
and marble 2 has a velocity of 3 m-s~^ in the positive x-direction. After they collide elastically, 
both marbles are moving. What is the final velocity of each marble? 

Exercise 6.10: Colliding Billiard Balls (Solution on p. 166.) 

Two billiard balls each with a mass of 150^ collide head-on in an elastic collision. Ball 1 was 
travelling at a speed of 2m • s~^ and ball 2 at a speed of 1,5m • s~^. After the collision, ball 1 
travels away from ball 2 at a velocity of 1,5m- s~^. 

1. Calculate the velocity of ball 2 after the collision. 

2. Prove that the collision was elastic. Show calculations. 



6.3.1.2 Inelastic Collisions 

Definition 6.2: Inelastic Collisions 

An inelastic collision is a collision in which total momentum is conserved but total kinetic 

energy is not conserved. The kinetic energy is transformed into other kinds of energy. 

So the total momentum before an inelastic collisions is the same as after the collision. But the total 
kinetic energy before and after the inelastic collision is different. Of course this does not mean that total 
energy has not been conserved, rather the energy has been transformed into another type of energy. 

As a rule of thumb, inelastic collisions happen when the colliding objects are distorted in some way. 
Usually they change their shape. The modification of the shape of an object requires energy and this is 
where the "missing" kinetic energy goes. A classic example of an inelastic collision is a motor car accident. 
The cars change shape and there is a noticeable change in the kinetic energy of the cars before and after 
the collision. This energy was used to bend the metal and deform the cars. Another example of an inelastic 
collision is shown in Figure 6.13. 



142 CHAPTER 6. MOTION IN TWO DIMENSIONS 



Image notjinished 

Figure 6.13: Asteroid moving towards the Moon. 



An asteroid is moving through space towards the Moon. Before the asteroid crashes into the Moon, the 
total momentum of the system is: 

Pi=Pim+Pia (6-17) 

The total kinetic energy of the system is: 

KE, = KE.ra + KE,a (6.18) 

When the asteroid collides inelastically with the Moon, its kinetic energy is transformed mostly into heat 
energy. If this heat energy is large enough, it can cause the asteroid and the area of the Moon's surface that 
it hits, to melt into liquid rock! From the force of impact of the asteroid, the molten rock flows outwards to 
form a crater on the Moon. 

After the collision, the total momentum of the system will be the same as before. But since this collision 
is inelastic, (and you can see that a change in the shape of objects has taken place!), total kinetic energy is 
not the same as before the collision. 

Momentum is conserved: 

Pi = Pf (6.19) 

But the total kinetic energy of the system is not conserved: 

KEi 7^ KEf (6.20) 

Exercise 6.11: An Inelastic Collision (Solution on p. 167.) 

Consider the collision of two cars. Car 1 is at rest and Car 2 is moving at a speed of 2 m-s~^ to 
the left. Both cars each have a mass of 500 kg. The cars collide inelastically and stick together. 
What is the resulting velocity of the resulting mass of metal? 

Exercise 6.12: Colliding balls of clay (Solution on p. 167.) 

Two balls of clay, 200g each, are thrown towards each other according to the following diagram. 
When they collide, they stick together and move off together. All motion is taking place in the 
horizontal plane. Determine the velocity of the clay after the collision. 



Image notjinished 

Figure 6.14 



143 

6.3.1.2.1 Collisions 

1. A truck of mass 4500 kg travelling at 20 m-s~^hits a car from behind. The car (mass 1000 kg) was 
travelling at 15 m-s~^. The two vehicles, now connected carry on moving in the same direction. 

a. Calculate the final velocity of the truck-car combination after the collision. 

b. Determine the kinetic energy of the system before and after the collision. 

c. Explain the difference in your answers for b). 

d. Was this an example of an elastic or inelastic collision? Give reasons for your answer. 

2. Two cars of mass 900 kg each collide and stick together at an angle of 90°. Determine the final velocity 
of the cars if car 1 was travelling at 15m-s~^and car 2 was travelling at 20m-s~^. 



Image notjinished 



Figure 6.15 



6.3.1.2.2 Tiny, Violent Collisions 

Author: Thomas D. Gutierrez 

Tom Gutierrez received his Bachelor of Science and Master degrees in Physics from San Jose State Univer- 
sity in his home town of San Jose, California. As a Master's student he helped work on a laser spectrometer 
at NASA Ames Research Centre. The instrument measured the ratio of different isotopes of carbon in CO2 
gas and could be used for such diverse applications as medical diagnostics and space exploration. Later, he 
received his PhD in physics from the University of California, Davis where he performed calculations for 
various reactions in high energy physics collisions. He currently lives in Berkeley, California where he studies 
proton-proton collisions seen at the STAR experiment at Brookhaven National Laboratory on Long Island, 
New York. 

High Energy Collisions 

Take an orange and expand it to the size of the earth. The atoms of the earth-sized orange would 
themselves be about the size of regular oranges and would fill the entire "earth-orange". Now, take an atom 
and expand it to the size of a football field. The nucleus of that atom would be about the size of a tiny 
seed in the middle of the field. From this analogy, you can see that atomic nuclei are very small objects by 
human standards. They are roughly 10~^^ meters in diameter - one-hundred thousand times smaller than 
a typical atom. These nuclei cannot be seen or studied via any conventional means such as the naked eye or 
microscopes. So how do scientists study the structure of very small objects like atomic nuclei? 

The simplest nucleus, that of hydrogen, is called the proton. Faced with the inability to isolate a single 
proton, open it up, and directly examine what is inside, scientists must resort to a brute-force and somewhat 
indirect means of exploration: high energy collisions. By colliding protons with other particles (such as other 
protons or electrons) at very high energies, one hopes to learn about what they are made of and how they 
work. The American physicist Richard Feynman once compared this process to slamming delicate watches 
together and figuring out how they work by only examining the broken debris. While this analogy may seem 
pessimistic, with sufficient mathematical models and experimental precision, considerable information can 
be extracted from the debris of such high energy subatomic collisions. One can learn about both the nature 
of the forces at work and also about the sub-structure of such systems. 

The experiments are in the category of "high energy physics" (also known as "subatomic" physics). The 
primary tool of scientific exploration in these experiments is an extremely violent collision between two very, 
very small subatomic objects such as nuclei. As a general rule, the higher the energy of the collisions, the 
more detail of the original system you are able to resolve. These experiments are operated at laboratories 



144 CHAPTER 6. MOTION IN TWO DIMENSIONS 

such as CERN, SLAC, BNL, and Fermilab, just to name a few. The giant machines that perform the 
collisions are roughly the size of towns. For example, the RHIC collider at BNL is a ring about 1 km in 
diameter and can be seen from space. The newest machine currently being built, the LHC at CERN, is a 
ring 9 km in diameter! 

6.3.1.2.3 Casestudy : Atoms and its Constituents 
Questions: 

1. What are isotopes? (2) 

2. What are atoms made up of? (3) 

3. Why do you think protons are used in the experiments and not atoms like carbon? (2) 

4. Why do you think it is necessary to find out what atoms are made up of and how they behave during 
collisions? (2) 

5. Two protons (mass 1,67 x 10^^^ kg) collide and somehow stick together after the collision. If each 
proton travelled with an initial velocity of 5, 00 x 10^ m- s~^ and they collided at an angle of 90°, what 
is the velocity of the combination after the collision. (9) 



6.4 Frames of reference* 
6.4.1 Frames of Reference 

6.4.1.1 Introduction 

Image notjinished 

Figure 6.16: Top view of a road with two people standing on opposite sides. A car drives past. 



Consider two people standing, facing each other on either side of a road. A car drives past them, heading 
West. For the person facing South, the car was moving toward the right. However, for the person facing 
North, the car was moving toward the left. This discrepancy is due to the fact that the two people used 
two different frames of reference from which to investigate this system. If each person were asked in what 
direction the car were moving, they would give a different answer. The answer would be relative to their 
frame of reference. 

6.4.1.2 What is a frame of reference? 

Definition 6.3: Frame of Reference 

A frame of reference is the point of view from which a system is observed. 

In practical terms, a frame of reference is a set of axes (specifying directions) with an origin. An observer 
can then measure the position and motion of all points in a system, as well as the orientation of objects in 
the system relative to the frame of reference. 

There are two types of reference frames: inertial and non-inertial. An inertial frame of reference travels 
at a constant velocity, which means that Newton's first law (inertia) holds true. A non-inertial frame of 



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145 

reference, such as a moving car or a rotating carousel, accelerates. Therefore, Newton's first law does not 
hold true in a non-inertial reference frame, as objects appear to accelerate without the appropriate forces. 

6.4.1.3 Why are frames of reference important? 

Frames of reference are important because (as we have seen in the introductory example) the velocity of a 
car can differ depending on which frame of reference is used. 

6.4.1.3.1 Frames of Reference and Special Relativity 

Frames of reference are especially important in special relativity, because when a frame of reference is moving 
at some significant fraction of the speed of light, then the flow of time in that frame does not necessarily 
apply in another reference frame. The speed of light is considered to be the only true constant between 
moving frames of reference. 

The next worked example will explain this. 

6.4.1.4 Relative Velocity 

The velocity of an object is frame dependent. More specifically, the perceived velocity of an object depends 
on the velocity of the observer. For example, a person standing on shore would observe the velocity of a 
boat to be different than a passenger on the boat. 

Exercise 6.13: Relative Velocity (Solution on p. 168.) 

The speedometer of a motor boat reads 5 m-s~^. The boat is moving East across a river which 
has a current traveling 3 m-s~^ North. What would the velocity of the motor boat be according to 
an observer on the shore? 

Exercise 6.14: Relative Velocity 2 (Solution on p. 169.) 

It takes a man 10 seconds to ride down an escalator. It takes the same man 15 s to walk back up 
the escalator against its motion. How long will it take the man to walk down the escalator at the 
same rate he was walking before? 



6.4.1.4.1 Frames of Reference 

1. A woman walks north at 3 km-hr~^ on a boat that is moving east at 4 km-hr~^. This situation is 
illustrated in the diagram below. 

a. How fast is the woman moving according to her friend who is also on the boat? 

b. What is the woman's velocity according to an observer watching from the river bank? 



Image notjinished 

Figure 6.17 



2. A boy is standing inside a train that is moving at 10 m-s ^to the left. The boy throws a ball in the 
air with a velocity of 4 m-s~^. What is the resultant velocity of the ball 

a. according to the boy? 

b. according to someone outside the train? 



146 CHAPTER 6. MOTION IN TWO DIMENSIONS 

This media object is a Flash object. Please view or download it at 

<http://static.slidesharecdn.com/swf/ssplayer2.swf?doc=gl2momentum-100513091859- 

phpapp01&stripped_title=gl2-momentum&userName=kwarne> 



Figure 6.18 



6.4,2 Summary 

1. Projectiles are objects that move through the air. 

2. Objects that move up and down (vertical projectiles) on the earth accelerate with a constant acceler- 
ation g which is approximately equal to 9,8 m-s~^directed downwards towards the centre of the earth. 

3. The equations of motion can be used to solve vertical projectile problems. 

Vf = Vi+ gt 

Ax = tellt ^ ^ 

Ax = Vit + \gt^ 

v"} = vf + 2gAx 



4. Graphs can be drawn for vertical projectile motion and are similar to the graphs for motion at constant 
acceleration. If upwards is taken as positive the Ax vs t, v vs t ans a vs t graphs for an object being 
thrown upwards look like this: 



Image notjinished 

Figure 6.19 



5. Momentum is conserved in one and two dimensions. 



p = 


mv 


Ap = 


- niAv 


Ap = 


-- FAt 



(6.22) 



6. An elastic collision is a collision where both momentum and kinetic energy is conserved. 

Pbefore ^ Paftcr , x 

KETocforc = KEaftcr 

7. An inelastic collision is where momentum is conserved but kinetic energy is not conserved. 

Pbcforc Paftcr 



(6.24) 

KE\^ciorc 7^ KEaftcr 



147 
8. The frame of reference is the point of view from which a system is observed. 

6.4,3 End of chapter exercises 

lEB 2005/11 HG Two friends, Ann and Lindiwe decide to race each other by swimming across a river to the other side. 
They swim at identical speeds relative to the water. The river has a current flowing to the east. 



Image notjinished 

Figure 6.20 



Ann heads a little west of north so that she reaches the other side directly across from the starting 
point. Lindiwe heads north but is carried downstream, reaching the other side downstream of Ann. 
Who wins the race? 

a. Ann 

b. Lindiwe 

c. It is a dead heat 

d. One cannot decide without knowing the velocity of the current. 

SC 2001/11 HGl A bullet fired vertically upwards reaches a maximum height and falls back to the ground. 



148 



CHAPTER 6. MOTION IN TWO DIMENSIONS 



r^ 



Figure 6.21 



Which one of the following statements is true with reference to the acceleration of the bullet during 
its motion, if air resistance is ignored? The acceleration: 

a. is always downwards 

b. is first upwards and then downwards 

c. is first downwards and then upwards 

d. decreases first and then increases 

SC 2002/03 HGl Thabo suspends a bag of tomatoes from a spring balance held vertically. The balance itself weighs 
10 N and he notes that the balance reads 50 N. He then lets go of the balance and the balance and 
tomatoes fall freely. What would the reading be on the balance while falling? 



Image notjinished 



Figure 6.22 



a. 50 N 

b. 40 N 

c. 10 N 

d. ON 



149 

lEB 2002/11 HGl Two balls, P and Q, are simultaneously thrown into the air from the same height above the ground. 
P is thrown vertically upwards and Q vertically downwards with the same initial speed. Which of 
the following is true of both balls just before they hit the ground? (Ignore any air resistance. Take 
downwards as the positive direction.) 





Velocity 


Acceleration 


A 


The same 


The same 


B 


P has a greater velocity than Q 


P has a negative acceleration; Q has a positive acceleration 


C 


P has a greater velocity than Q 


The same 


D 


The same 


P has a negative acceleration; Q has a positive acceleration 



Table 6.1 

lEB 2002/11 HGl An observer on the ground looks up to see a bird flying overhead along a straight line on bearing 130° 
(40° S of E). There is a steady wind blowing from east to west. In the vector diagrams below, I, II and 
III represent the following: I the velocity of the bird relative to the air II the velocity of the 

air relative to the ground III the resultant velocity of the bird relative to the ground Which diagram 
correctly shows these three velocities? 



Image notjtnished 



Figure 6.23 



SC 2003/11 A ball X of mass m is projected vertically upwards at a speed Ux from a bridge 20 m high. A ball Y 
of mass 2m is projected vertically downwards from the same bridge at a speed of Uy. The two balls 
reach the water at the same speed. Air friction can be ignored. Which of the following is true with 
reference to the speeds with which the balls are projected? 



a. 
b. 
c. 
d. 



2"-l 

Uy 
2Uy 

4u„ 



SC 2002/03 HGl A stone falls freely from rest from a certain height. Which one of the following quantities could be 
represented on the y-axis of the graph below? 



Image notjinished 



Figure 6.24 



a. velocity 

b. acceleration 

c. momentum 

d. displacement 



150 



CHAPTER 6. MOTION IN TWO DIMENSIONS 



1. A man walks towards the back of a train at 2 m-s ^while the train moves forward at 10 m-s ^. The 
magnitude of the man's velocity with respect to the ground is 

a. 2 m-s~^ 

b. 8 m-s~^ 

c. 10 m-s~^ 

d. 12 m-s-i 

2. A stone is thrown vertically upwards and it returns to the ground. If friction is ignored, its acceleration 
as it reaches the highest point of its motion is 

a. greater than just after it left the throwers hand. 

b. less than just before it hits the ground. 

c. the same as when it left the throwers hand. 

d. less than it will be when it strikes the ground. 

3. An exploding device is thrown vertically upwards. As it reaches its highest point, it explodes and 
breaks up into three pieces of equal mass. Which one of the following combinations is possible for 
the motion of the three pieces if they all move in a vertical line? 





Mass 1 


Mass 2 


Mass 3 


A 


V downwards 


V downwards 


V upwards 


B 


V upwards 


2v downwards 


V upwards 


C 


2v upwards 


V downwards 


V upwards 


D 


V upwards 


2v downwards 


V downwards 



Table 6.2 



lEB 2004/11 HGl A stone is thrown vertically up into the air. Which of the following graphs best shows the resultant 
force exerted on the stone against time while it is in the air? (Air resistance is negligible.) 



Image notjinished 



Figure 6.25 



4. What is the velocity of a ball just as it hits the ground if it is thrown upward at 10 m-s~^from a height 
5 meters above the ground? 
lEB 2005/11 HGl A breeze of 50 km-hr"'^ blows towards the west as a pilot flies his light plane from town A to village B. 
The trip from A to B takes 1 h. He then turns west, flying for ^ h until he reaches a dam at point C. 
He turns over the dam and returns to town A. The diagram shows his flight plan. It is not to scale. 



Image notjinished 



Figure 6.26 



The pilot flies at the same altitude at a constant speed of 130 km.h ^ relative to the air throughout 
this flight. 



151 

a. Determine the magnitude of the pilot's resultant velocity from the town A to the village B. 

b. How far is village B from town A? 

c. What is the plane's speed relative to the ground as it travels from village B to the dam at C? 

d. Determine the following, by calculation or by scale drawing: 

1. The distance from the village B to the dam C. 

2. The displacement from the dam C back home to town A. 

5. A cannon (assumed to be at ground level) is fired off a fiat surface at an angle, above the horizontal 
with an initial speed of fo- 

a. What is the initial horizontal component of the velocity? 

b. What is the initial vertical component of the velocity? 

c. What is the horizontal component of the velocity at the highest point of the trajectory? 

d. What is the vertical component of the velocity at that point? 

e. What is the horizontal component of the velocity when the projectile lands? 

f. What is the vertical component of the velocity when it lands? 

lEB 2004/11 HGl Hailstones fall vertically on the hood of a car parked on a horizontal stretch of road. The average 
terminal velocity of the hailstones as they descend is 8,0 m.s~^ and each has a mass of 1,2 g. 

a. Calculate the magnitude of the momentum of a hailstone just before it strikes the hood of the 
car. 

b. If a hailstone rebounds at 6,0 m.s~^ after hitting the car's hood, what is the magnitude of its 
change in momentum? 

c. The hailstone is in contact with the car's hood for 0,002 s during its collision with the hood of the 
car. What is the magnitude of the resultant force exerted on the hood if the hailstone rebounds 
at 6,0 m.s~^? 

d. A car's hood can withstand a maximum impulse of 0,48 N-s without leaving a permanent dent. 
Calculate the minimum mass of a hailstone that will leave a dent in the hood of the car, if it falls 
at 8,0 m.s~^ and rebounds at 6,0 m.s~^ after a collision lasting 0,002 s. 

3/11 HGl - Biathlon Andrew takes part in a biathlon race in which he first swims across a river and then cycles. The 
diagram below shows his points of entry and exit from the river, A and P, respectively. 



Image notjinished 



Figure 6.27 



During the swim, Andrew maintains a constant velocity of 1,5 m.s~^ East relative to the water. The 
water in the river flows at a constant velocity of 2,5 m.s~^ in a direction 30° North of East. The width 
of the river is 100 m. The diagram below is a velocity- vector diagram used to determine the resultant 
velocity of Andrew relative to the river bed. 



Image notjinished 

Figure 6.28 

a. Which of the vectors (AB, BC and AC) refer to each of the following? 



152 CHAPTER 6. MOTION IN TWO DIMENSIONS 

1. The velocity of Andrew relative to the water. 

2. The velocity of the water relative to the water bed. 

3. The resultant velocity of Andrew relative to the river bed. 

b. Determine the magnitude of Andrew's velocity relative to the river bed either by calculations or 
by scale drawing, showing your method clearly. 

c. How long (in seconds) did it take Andrew to cross the river? 

d. At what distance along the river bank (QP) should Peter wait with Andrew's bicycle ready for 
the next stage of the race? 

HGl - Bouncing Ball A ball bounces vertically on a hard surface after being thrown vertically up into the air by a boy 
standing on the ledge of a building. Just before the ball hits the ground for the first time, it has a 
velocity of magnitude 15 m.s~^. Immediately, after bouncing, it has a velocity of magnitude 10 m.s~^. 
The graph below shows the velocity of the ball as a function of time from the moment it is thrown 
upwards into the air until it reaches its maximum height after bouncing once. 



Image notjinished 

Figure 6.29 



a. At what velocity does the boy throw the ball into the air? 

b. What can be determined by calculating the gradient of the graph during the first two seconds? 

c. Determine the gradient of the graph over the first two seconds. State its units. 

d. How far below the boy's hand does the ball hit the ground? 

e. Use an equation of motion to calculate how long it takes, from the time the ball was thrown, for 
the ball to reach its max:imum height after bouncing. 

f. What is the position of the ball, measured from the boy's hand, when it reaches its maximum 
height after bouncing? 

lEB 2001/11 HGl - Free Falling? A parachutist steps out of an aircraft, flying high above the ground. She falls for the 
first 8 seconds before opening her parachute. A graph of her velocity is shown in Graph A below. 



Image notjinished 

Figure 6.30 



Use the information from the graph to calculate an approximate height of the aircraft when she 
stepped out of it (to the nearest 10 m). 

What is the magnitude of her velocity during her descent with the parachute fully open? The 
air resistance acting on the parachute is related to the speed at which the parachutist descends. 
Graph B shows the relationship between air resistance and velocity of the parachutist descending 
with the parachute open. 



153 

Image notjinished 

Figure 6.31 



c. Use Graph B to find the magnitude of the air resistance on her parachute when she was descending 
with the parachute open. 

d. Assume that the mass of the parachute is negligible. Calculate the mass of the parachutist showing 
your reasoning clearly. 

6. An aeroplane travels from Cape Town and the pilot must reach Johannesburg, which is situated 1300 km 
from Cape Town on a bearing of 50° in 5 hours. At the height at which the plane files, a wind is blowing 
at 100 km-hr~^on a bearing of 130 ° for the whole trip. 



Image notjinished 

Figure 6.32 



a. Calculate the magnitude of the average resultant velocity of the aeroplane, in km-hr~^, if it is to 
reach its destination on time. 

b. Calculate ther average velocity, in km-hr~^, in which the aeroplane should be travelling in order 
to reach Johannesburg in the prescribed 5 hours. Include a labelled, rough vector diagram in your 
answer. (If an accurate scale drawing is used, a scale of 25 km-hr~^= 1 cm must be used.) 

7. Niko, in the basket of a hot-air balloon, is stationary at a height of 10 m above the level from where 
his friend, Bongi, will throw a ball. Bongi intends throwing the ball upwards and Niko, in the basket, 
needs to descend (move downwards) to catch the ball at its maximum height. 



Image notjinished 

Figure 6.33 



Bongi throws the ball upwards with a velocity of 13 m-s~^. Niko starts his descent at the same instant 
the ball is thrown upwards, by letting air escape from the balloon, causing it to accelerate downwards. 
Ignore the effect of air friction on the ball. 

a. Calculate the maximum height reached by the ball. 

b. Calculate the magnitude of the minimum average acceleration the balloon must have in order for 
Niko to catch the ball, if it takes 1,3 s for the ball to rach its maximum height. 

8. Lesedi (mass 50 kg) sits on a massless trolley. The trolley is travelling at a constant speed of 3 m-s~^. 
His friend Zola (mass 60 kg) jumps on the trolley with a velocity of 2 m-s~^. What is the final velocity 
of the combination (Lesedi, Zola and trolley) if Zola jumps on the trolley from 

a. the front 

b. behind 



154 CHAPTER 6. MOTION IN TWO DIMENSIONS 

c. the side 
(Ignore all kinds of friction) 

Image not finished 

Figure 6.34 



155 

Solutions to Exercises in Chapter 6 

Solution to Exercise 6.1 (p. 135) 

Step 1. We are required to determine the maximum height reached by the ball and how long it takes to reach 
this height. We are given the initial velocity Vi = 10 m-s~^and the acceleration due to gravity g = 9,8 
m-s~^. 

Step 2. Choose down as positive. We know that at the maximum height the velocity of the ball is m-s~^. 
We therefore have the following: 

• Vi = —10 m ■ s^^ {it IS negative because we chose downwards as positive) 

• Vf = Om ■ s^^ 

• g = +9, 8 m • s"^ 

Step 3. We can use: 

(6.25) 









vj = vf + 2grAa; 




to solve for the height. 






Step 4. 












-} 


= vf + 2gl\x 






{Of 


= (-10)V(2)(9,8)(Ax) 






-100 


19,6Aa; 






Ax 


= -5,102m 



(6.26) 



The value for the displacement will be negative because the displacement is upwards and we have chosen 
downward as positive (and upward as negative). The height will be a positive number, h = 5, 10m. 
Step 5. We can use: 

Vf = v,+ gt (6.27) 

to solve for the time. 
Step 6. 

Vf = Vi+ gt 

= -10 + 9,8t 

(6.28) 
10 = 9,8^ 

t = 1,02s 

Step 7. The ball reaches a maximum height of 5,10 m. 
The ball takes 1,02 s to reach the top. 

Solution to Exercise 6.2 (p. 135) 

Step 1. We need to find how high the ball goes. We know that it takes 10 seconds to go up and down. We do 
not know what the initial velocity of the ball {vi) is. 

Step 2. A problem like this one can be looked at as if there are two parts to the motion. The first is the ball 
going up with an initial velocity and stopping at the top (final velocity is zero). The second motion is 
the ball falling, its initial velocity is zero and its final velocity is unknown. 



Image notjinished 

Figure 6.35 



156 CHAPTER 6. MOTION IN TWO DIMENSIONS 

Choose down as positive. We know that at the maximum height, the velocity of the ball is m-s~^. 
We also know that the ball takes the same time to reach its maximum height as it takes to travel from 
its maximum height to the ground. This time is half the total time. We therefore know the following 
for the second motion of the ball going down: 

• t = 5 s, half of the total time 

• vtop = Vi = m ■ s~'^ 

• g = 9,8m ■ s^^ 

• Aa;= ? 

Step 3. We do not know the final velocity of the ball coming down. We need to choose an equation that does 
not have Vf in it. We can use the following equation to solve for Ax: 

(6.29) 
Step 4. 

(6.30) 

In the second motion, the displacement of the ball is 122,5m downwards. This means that the height 
was 122,5m, h=122,5m. 
Step 5. The ball reaches a maximum height of 122,5 m. 

Solution to Exercise 6.3 (p. 136) 

Step 1. We are required to draw graphs of 

a. Ax vs. t 

b. V vs. t 

c. a vs. t 

Step 2. There are two parts to the motion of the ball: 

a. ball travelling upwards from the building 

b. ball falling to the ground 

We examine each of these parts separately. To be able to draw the graphs, we need to determine the 
time taken and displacement for each of the motions. 
Step 3. For the first part of the motion we have: 

• Vi = +4, 9 m ■ s~^ 





Ax = v,t + -gt^ 


Ax 


= (0)(5)+i(9,8)(5) 


Ax 


0+122, 5m 



• 



Vf = Qm-s ^ 
• g = —9, 8m • s 



-2 



Iviiage notjinished 

Figure 6.36 



157 
Therefore we can use v'j = vf + 2gAx to solve for the height and Vf = Vi + gt to solve for the time. 



(6.31) 



vj = 


vf + 2gAx 


{of = 


(4,9)^2 X (-9,8) xAx 


',6Aa; = 


(4,9)' 


Ax = 


1,225 m 



Vf = Vi + gt 

= 4,9+ (-9,8) xf 

9,8t = 4,9 

t = 0,5 s 



Image notjinished 

Figure 6.37 



Step 4. For the second part of the motion we have: 

• Vi = Om ■ s~^ 

• Aa;= -(20+ 1,225) m 

• g = —9, 8m • s^^ 

Therefore we can use Ax = vit + \gt^ to solve for the time. 



Ax 


= 




v,t + \ge 


(20+1,225) 


= 


(0) xt+i X (-9,8) xe 


-21,225 


= 




0-4, 9^2 


e 


= 




4,33163... 


t 


= 




2,08125... s 



(6.32) 



(6.33) 



Image notjinished 

Figure 6.38 



Step 5. The ball starts from a position of 20 m (at t = s) from the ground and moves upwards until it reaches 
(20 + 1,225) m (at t = 0,5 s). It then falls back to 20 m (at t = 0,5 + 0,5 = 1,0 s) and then falls to 
the ground, A x = m at (t = 0,5 + 2,08 = 2,58 s). 



158 CHAPTER 6. MOTION IN TWO DIMENSIONS 



Image notjinished 



Figure 6.39 



Step 6. The ball starts off with a velocity of +4,9 m-s~^at t = s, it then reaches a velocity of m-s~^at t 
= 0,5 s. It stops and falls back to the Earth. At t = 1,0s (i.e. after a further 0,5s) it has a velocity 
of -4,9 m-s~^. This is the same as the initial upwards velocity but it is downwards. It carries on at 
constant acceleration until t = 2,58 s. In other words, the velocity graph will be a straight line. The 
final velocity of the ball can be calculated as follows: 

Vf = Vi + gt 

= 0+ (-9, 8) (2, 08...) (6.34) 

= -20, 396... m-s-i 



Image notjinished 

Figure 6.40 



Step 7. We chose upwards to be positive. The acceleration of the ball is downward, g = —9.8 m • s 
the acceleration is constant throughout the motion, the graph looks like this: 



Image notjinished 



Figure 6.41 



Solution to Exercise 6.4 (p. 136) 

Step 1. The initial position of the ball will tell us how high the window is. From the y-axis on the graph we 

can see that the ball is 4 m from the ground. 

The window is therefore 4 m above the ground. 
Step 2. The maximum height is where the position-time graph show the maximum position - the top of the 

curve. This is when t = 0,2 s. 

It takes the ball 0,2s to reach the maximum height. 
Step 3. To find the initial velocity we only look at the first part of the motion of the ball. That is from when 

the ball is released until it reaches its maximum height. We have the following for this: In this case, 

let's choose upwards as positive. 

t = 0,2 s 

g = -9,8 m-s-2 (6.35) 

Vf = m ■ s^^ (because the ball stops) 



Vf = 


Vi + gt 


= 


i;, + (-9,8)(0,2) 


Vt = 


1,96m -s-i 



159 

To calculate the initial velocity of the ball {vi), we use: 

gt 

(6.36) 



The initial velocity of the ball is 1,96 m-s ^upwards. 
Step 4. To find the maximum height we look at the initial motion of the ball. We have the following: 

t = 0,2 s 

g = —9,8 m • s^^ 

Vf = m ■ s~^ (because the ball stops) 

Vi = +1, 96 m • s^^ (calculated above) 

To calculate the displacement from the window to the maximum height (Ax) we use: 



(6.37) 



Ax = Vit + \gt^ 

Ax = (1,96) (0,2) +i (-9,8) (0,2)2 (g_3g) 

Ax = 0,196m 

The maximum height of the ball is (4 + 0,196) = 4,196 m above the ground. 
Step 5. To find the final velocity of the ball we look at the second part of the motion. For this we have: 

Ax = —4, 196 m (because upwards is positive) 
g = -9,8 m-s-2 (6.39) 

Vi = m ■ s~^ 

We can use {vf) = {vi) + 2gAx to calculate the final velocity of the ball. 



(6.40) 



The final velocity of the ball is 9,07 m-s~^ downwards. 

Solution to Exercise 6.5 (p. 136) 

Step 1. We need to study the velocity-time graph to answer this question. We will break the motion of the 
ball up into two time zones: t = 0stot = 2s and t = 2stot = 4s. 
From t = 0stot = 2s the following happens: 

The ball starts to move at an initial velocity of 19,6 m-s~^and decreases its velocity to m-s~^at t = 
2 s. At t = 2 s the velocity of the ball is m-s~^and therefore it stops. 
From t = 2stot = 4s the following happens: 

The ball moves from a velocity of m-s~^to 19,6 m-s~^in the opposite direction to the original motion. 
If we assume that the ball is hit straight up in the air (and we take upwards as positive), it reaches its 
maximum height at t = 2 s, stops, turns around and falls back to the Earth to reach the ground at t 
= 4 s. 



{Vff = 


{vif + 2gAx 


i^ff = 


(0)^2 (-9, 8) (-4, 196) 


(Vff = 


82,2416 


Vf = 


9,0687...m-s"i 



160 CHAPTER 6. MOTION IN TWO DIMENSIONS 

Step 2. To draw this graph, we need to determine the displacements at t = 2 s and t = 4 s. 
At t = 2 s: 

The displacement is equal to the area under the graph: 
Area under graph = Area of triangle 
Area = ^bh 
Area = ^x 2 x 19,6 
Displacement = 19,6 m 
At t = 4 s: 

The displacement is equal to the area under the whole graph (top and bottom). Remember that an 
area under the time line must be substracted: 
Area under graph = Area of triangle 1 + Area of triangle 2 
Area = ^bh + ^bh 

Area = (^x 2 x 19,6) + (ix 2 x (-19,6)) 
Area = 19,6 - 19,6 
Displacement = m 

The displacement-time graph for motion at constant acceleration is a curve. The graph will look like 
this: 



Image notjinished 

Figure 6.42 



Step 3. To draw the acceleration vs. time graph, we need to know what the acceleration is. The velocity-time 
graph is a straight line which means that the acceleration is constant. The gradient of the line will 
give the acceleration. 

The line has a negative slope (goes down towards the left) which means that the acceleration has a 
negative value. 

Calculate the gradient of the line: 
gradient = ^ 
gradient = 5=^ 
gradient = ~ ^ ' 
gradient = -9,8 
acceleration = 9,8 m-s~^downwards 



Image notjinished 

Figure 6.43 



Solution to Exercise 6.6 (p. 139) 

Step 1. The first step is to draw the picture to work out what the situation is. Mark the initial velocities of 
both players in the picture. 



161 



Image notjinished 



Figure 6.44 



We also know that rrii = 1112 = 90 kg and Vf2 = ms~^. 

We need to find the final velocity and angle at which Player 1 bounces off Player 2. 
Step 2. Total initial nionientuni = Total final nionientuni. But we have a two dimensional problem, and we 
need to break up the initial momentum into x and y components. 



For Player 1: 



For Player 2: 



Step 3. For Player 1: 



For Player 2: 



Pix — Pfx 
Piy = Pfy 

Pixi = rniViix = 90 X = 
Piyi = niiViiy = 90 X 5 

Pix2 = fn2Vi2x = 90 X 8 X sin&{f 
Piy2 = 'fn2Vi2y = 90 X 8 X cos60° 

Pfxi = miVfxi = 90 X Vfxi 
Pfyi = '^I'Vfyi = 90 X Vfyi 

Pfx2 = n^2«/x2 = 90 X = 
Pfy2 = 'm2Vfy2 = 90 X = 



(6.41) 



(6.42) 



(6.43) 



(6.44) 



(6.45) 



Step 4. The initial total momentum in the x direction is equal to the final total momentum in the x direction. 
The initial total momentum in the y direction is equal to the final total momentum in the y direction. 
If we find the final x and y components, then we can find the final total momentum. 

Ptxl + Pix2 = Pfxl + Pfx2 

+ 90 X 8 X .sm60° = 90 x Vf^i + 



Vfxl 



(6-46) 



90 

^/a;i = 6.928ms~^ 

Ptyl+Pty2 = Pfyl+Pfy2 

90 X 5 + 90 X 8 X cos60° = 90 x Vfyi + 



(6-47) 

,, _ 90x5+90x8xeos60° ^ ' 

fv^ ~ 90 

Vfyi = 9.0ms~^ 



162 CHAPTER 6. MOTION IN TWO DIMENSIONS 

Step 5. Use Pythagoras's theorem to find the total final velocity: 

Image notjinished 

Figure 6.45 



^ftot = JV%,+V%, 



v/6.9282 + 92 (6.48) 

11.36 



Calculate the angle 6 to find the direction of Player I's final velocity: 

sine = ^^^^ 

■"ftot 



6 = 52.4= 



(6.49) 



Therefore Player 1 bounces off Player 2 with a final velocity of 11.36 m-s ^ at an angle of 52.4° from 
the horizontal. 

Solution to Exercise 6.7 (p. 139) 

Step 1. The first step is to draw the picture to work out what the situation is. Mark the initial velocities of 
both players in the picture. 



Image notjinished 

Figure 6.46 



We need to define a reference frame: For y, choose the direction they are both running in as positive. 
For X, the direction player 2 is running in is positive. 

We also know that mi = 1112 = 80 kg. And w/a;2=-0.3 ms~^ and w/j,2=6 ms~^. 
We need to find the final velocity and angle at which Player 1 bounces off Player 2. 
Step 2. Total initial momentum = Total final momentum. But we have a two dimensional problem, and we 
need to break up the initial momentum into x and y components. Remember that momentum is a 
vector and has direction which we will indicate with a '+' or '-' sign. 

^" = ^^" (6.50) 

Piy = Pfy 

For Player 1: 

Pixi = TOiWiix = 80 X (-5) X cos75° 

(6.51) 
Piyi = rriiViiy = 80 X 5 X sm75 



163 



For Player 2: 



Step 3. For Player 1: 



For Player 2: 



Pix2 = 


m2Vi2x = 80 X 6 X cosQQ' 


Pty2 = 


"T'2^i2i/ = 80 X 6 X sm60' 


Pfxl 


= miVfxi = 80 X Vfxi 


Pfyi 


= miVfyl = 80 X Vfyl 



(6.52) 



(6.53) 



Pfx2 = m2Vf^2 = 80 X (-0.3) 

Pfy2 = ™2W/j/2 = 80 X 6 

Step 4. The initial total momentum in the x direction is equal to the final total momentum in the x direction. 
The initial total momentum in the y direction is equal to the final total momentum in the y direction. 
If we find the final x and y components, then we can find the final total momentum. 



Pixl+Pix2 = Pfxl+Pfx2 

-80 X 5cos75° + 80 X 6 X cos60° = 80 x w/a-i + 80 x (-0.3) 



Vfxl = 


- 


-80x5cos75' 


'+80x6xcos60°-80x(-0.3) 
80 


Vfxl = 








2.0 ms^i 


Piyl + Ptv2 


= 






Pfyi + Pfy2 


80 X 5 X sm75° + 80 x 6 x sm60° 


= 




80 


X Vfyl + 80 X 6 


«/yi 


= 


80x5 


sin' 


75°+80x6xsin60°-80x6 
80 


Vfyl 


= 






4.0 ms-1 



(6.55) 



(6.56) 



Step 5. Use Pythagoras's theorem to find the total final velocity: 

Image notjtnished 



Figure 6.47 



«/*«* = Vz-i + ^/yi 



V22 + 42 (6.57) 

4.5 



Calculate the angle 6 to find the direction of Player I's final velocity: 

tane = ^-^ 



63.4= 



(6.58) 



Therefore Player 1 bounces off Player 2 with a final velocity of 4.5 m-s ^ at an angle of 63.4° from 
the horizontal. 



164 CHAPTER 6. MOTION IN TWO DIMENSIONS 

Solution to Exercise 6.8 (p. 141) 

Step 1. We are given: 

• mass of ball 1, nii = 0.3 kg 

• mass of ball 2, m2 = 0.3 kg 

• initial velocity of ball 1, f^i = m-s~^ 

• initial velocity of ball 2, Vi2 = 2 m-s"-'^ 

• final velocity of ball 2, Vf2 = m-s~^ 

• the collision is elastic 

All quantities are in SI units. We are required to determine the final velocity of ball 1, Vfi. Since the 
collision is elastic, we know that 

• momentum is conserved, niiVn + m2Vi2 = rriiVfi + m2W/2 



• energy is conserved, ^ (^rriivf-^^ + m2vf2) = \ ( 
Step 2. Choose to the right as positive. 



mivj^ + m2vJ2 



Image notjinished 



Step 3. 

Figure 6.48 



Step 4. Momentum is conserved. Therefore: 



Pi = Pf 

miVii + m2Vi2 = miVfi + m2W/2 
(0,3)(0) + (0,3)(2) = (0,3)«/i + 



Vfi 

Step 5. The final velocity of ball 1 is 2 m-s~^to the right. 
Solution to Exercise 6.9 (p. 141) 

Step 1. We are given: 

• mass of marble 1, toi=50 g 

• mass of marble 2, m2=100 g 

• initial velocity of marble 1, Vii=0 m-s~^ 

• initial velocity of marble 2, Wi2=3 m-s~^ 

• the collision is elastic 

The masses need to be converted to SI units. 



TOi = 0,05 kg 
1712 = 0, 1 kg 



We are required to determine the final velocities: 

• final velocity of marble 1, Vfi 

• final velocity of marble 2, Vf2 

Since the collision is elastic, we know that 



(6.59) 



(6.60) 



165 

• momentum is conserved, pi = Pf- 

• energy is conserved, KEi=KEf. 

We have two equations and two unknowns (wi, V2) so it is a simple case of solving a set of simultaneous 
equations. 
Step 2. Choose to the right as positive. 



Itmige notjinished 



Step 3. 

Figure 6.49 



Step 4. Momentum is conserved. Therefore: 



Pi = Pf 

Pil+Pi2 = Pfl+Pf2 

miVii -\- m2Vi2 = niiVfi + m2Vf2 (6.61) 

(0, 05) (0) + (0,1) (3) = (0,05)i;/i + (0,l)i;/2 

0,3 = 0,05w/i + 0, li;/2 



f2 



Energy is also conserved. Therefore: 



KE, = 


KEf 


KEa + KE,2 = 


KEfi + KEf2 


Imivl + \m2V% = 


^mivj^ + lm2vj^ 



(i)(0,05)(0)V(i)(0,l)(3)^ = l{0,05){vf,f+{'^){0,l){vf2f 

0,45 = 0, 025i;^i + 0, 05w^2 

Substitute Equation (6.61) into Equation (6.62) and solve for Vf2- 

m2vJ2 = m-iW/i + m-2W/2 

mi(^(i;,2-«/2)) +m2v}^ 

2 , _ _,,2 



mi^(Wi2-W/2) +m2Wj2 

:^^{V^2-Vf2) +m2Vf^ 

^{V.2-Vf2f + V% 



Vf2 = ^{V,2-Vf2f + V' 



'Jf2 



^ {vf, - 2 . t;,2 • Vf2 + v%) + v) 

(uife-l)(3)'-2fM(3)-«/2+(c;^ + l)«?2 
(2 - 1) (3)^ - 2 • 2 (3) • Vf2 + (2 + 1) v% 

Vf2 + 3w^2 
Vf2 + VJ^ 
{Vf2 - 3) {Vf2 - 1) 



9- 12wf2 + 3w? 



3-4wf2 + w? 



(6.62) 



(6.63) 



166 CHAPTER 6. MOTION IN TWO DIMENSIONS 

Substituting back into Equation (6.61), we get: 



Vfl 


= 


^K2-M 




= 


cM(3-3) 




= 


m- s-'^ 




or 




Vfi 


= 


^K2-«/2) 




= 


<^(3-l) 




= 


4 m • s~^ 



(6.64) 



But according to the question, marble 1 is moving after the collision, therefore marble 1 moves to the 
right at 4 m-s~^. Substituting this value for Vfi into Equation (6.61), we can calculate: 

(6.65) 



Therefore marble 2 moves to the right with a velocity of 1 m-s ^. 
Solution to Exercise 6.10 (p. 141) 



0,3- 


-0,05t>/i 




0,1 


0,3 


-0,05x4 




0,1 


1 ', 


m ■ s^^ 



Image notjinished 



Step 1. a. 

Figure 6.50 

b. Since momentum is conserved in all kinds of collisions, we can use conservation of momentum to 

solve for the velocity of ball 2 after the collision. 
c. 

Pbefore — Pafter 

miVii -\- m2Vi2 = niiVfi + m2Vf2 

(S(2) + (S(-1,5) = S)(-l,5) + (l|)(«,2) (6.66) 

0,3-0,225 = -0,225 + 0,15w/2 

Vf2 = ^m ■ s~^ 

So after the collision, ball 2 moves with a velocity of 3?tt. • s~^. 
Step 2. The fact that characterises an elastic collision is that the total kinetic energy of the particles before 
the collision is the same as the total kinetic energy of the particles after the collision. This means that 
if we can show that the initial kinetic energy is equal to the final kinetic energy, we have shown that 
the collision is elastic. Calculating the initial total kinetic energy: 

EKbefore = l"^l«j^l + |"T-2Wj^2 

= (i)(0,15)(2)V(i)(0,15)(-l,5)' (6.67) 

0.469.... J 



167 



Calculating the final total kinetic energy: 



EKafter = S^^l^/l + 5"T'2W/2 

= (i)(0,15)(-l,5)V(i)(0,15)(2)2 (6.68) 

0.469... .J 



So EKbefore = EKafter and hence the collision is elastic. 
Solution to Exercise 6.11 (p. 142) 



Image notjinished 



Step 1. 

Figure 6.51 



Step 2. We are given: 

• mass of car 1, mi = 500 kg 

• mass of car 2, m2 = 500 kg 

• initial velocity of car 1, w^i = m-s~^ 

• initial velocity of car 2, Vi2 = 2 m-s~^to the left 

• the collision is inelastic 

All quantities are in SI units. We are required to determine the final velocity of the resulting mass, Vf. 
Since the collision is inelastic, we know that 

• momentum is conserved, miVn + m2Vi2 = rriiVfi + 7712^/2 

• kinetic energy is not conserved 

Step 3. Choose to the left as positive. 

Step 4. So we must use conservation of momentum to solve this problem. 



P^ 


- 


Pf 


Pil + Pi2 


= 


Pf 


miVii + m2Vi2 


= 


{nil + m2)vf 


(0) + (500) (2) 


= 


(500 + 500) Vf 


1000 


= 


WOOvf 


Vf 


= 


1 m ■ s^^ 



(6.69) 



Therefore, the final velocity of the resulting mass of cars is 1 m-s ^to the left. 
Solution to Exercise 6.12 (p. 142) 

Step 1. This is an inelastic collision where momentum is conserved. 

The momentum before = the momentum after. 

The momentum after can be calculated by drawing a vector diagram. 
Step 2. 

pi (before) = miVn = (0, 2) (3) = 0, 6kg • mcdots"^east 
P2 (before) = m2Vi2 = (0, 2) (4) = 0, 8kg • TTicdots^^south 



(6.70) 



168 CHAPTER 6. MOTION IN TWO DIMENSIONS 

Step 3. Here we need to draw a diagram: 

Image notjinished 

Figure 6.52 



Pi+2 (after) = y^(0, 8)' + (0, 6)' 
= 1 

Step 4. First we have to find the direction of the final momentum: 

tanO = TTTj 

0,6 

e = 53,13° 
Now we have to find the magnitude of the final velocity: 



Solution to Exercise 6.13 (p. 145) 

Image notjinished 



Step 1. 

Figure 6.53 



Step 2. 



5,8 m ■ s ^ 



tanO = I 

e = 59,04° 



Ittiage notjinished 

Figure 6.54 



(6.71) 



(6.72) 



Pl+2 = mi+2Vf 

1 = (0,2 + 0,2)1;/ (6.73) 

Vf = 2,5m- s^^ 53, 13°SouthofEast 



R = ^{3f + {5f 

VM (6.74) 



(6.75) 



169 

The observer on the shore sees the boat moving with a velocity of 5,8 m-s~^ at 59,04°east of north due 
to the current pushing the boat perpendicular to its velocity. This is contrary to the perspective of a 
passenger on the boat who perceives the velocity of the boat to be 5 m-s~^ due East. Both perspectives 
are correct as long as the frame of the observer is considered. 

Solution to Exercise 6.14 (p. 145) 

Step 1. We are required to determine the time taken for a man to walk down an escalator, with its motion. 

We are given the time taken for the man to ride down the escalator and the time taken for the man to 

walk up the escalator, against it motion. 
Step 2. Select down as positive and assume that the escalator moves at a velocity Ve- If the distance of the 

escalator is Xe then: 

Ve=^ (6.76) 

10s 

Now, assume that the man walks at a velocity «„. Then we have that: 

Ve-Vm= YT' (6.77) 



We are required to find t in: 



Ve + Vm = Y' (6.78) 



Step 3. We find that we have three equations and three unknowns {ve, Wm and t). 
Add (6.77) to (6.78) to get: 



15 s t 
Substitute from (6.76) to get: 



2ve = ^ + — (6.79) 



10s 15s t 
Since Xe is not equal to zero we can divide throughout by Xe- 



2^ = ^ + ? (6-80) 



Re- write: 



Multiply by t: 



Solve for t: 



2 _ 1 1 
TOs ~ T5s "^ i 

2 1 _ 1 

10^ ~ 15^ ~ 1 



_2 ]_ 

10 s 15 s 



to get: 



Step 4. The man will take 15/2 = 7,5 s 



(6.81) 



(6.82) 



*(t^-i^)=i (^-^^^ 



(6.84) 



15 

- s (6.85) 



170 CHAPTER 6. MOTION IN TWO DIMENSIONS 



Chapter 7 

Mechanical properties of matter 



7.1 Deformation of materials' 

7.1.1 Introduction 

In this chapter we will look at some mechanical (physical) properties of various materials that we use. The 
mechanical properties of a material are those properties that are affected by forces being applied to the 
material. These properties are important to consider when we are constructing buildings, structures or 
modes of transport like an aeroplane. 

7.1.2 Deformation of materials 

7.1.2.1 Hooke's Law 

Deformation (change of shape) of a solid is caused by a force that can either be compressive or tensile when 
applied in one direction (plane). Compressive forces try to compress the object (make it smaller or more 
compact) while tensile forces try to tear it apart. We can study these effects by looking at what happens 
when you compress or expand a spring. 

Hooke's Law relates the restoring force of a spring to its displacement from equilibrium length. 

The equilibrium length of a spring is its length when no forces are applied to it. When a force is applied 
to a spring, e.g., by attaching a weight to one end, the spring will expand and become longer. The difference 
between the new length and the equilibrium length is the displacement. 

NOTE: Hooke's law is named after the seventeenth century physicist Robert Hooke who discovered 
it in 1660 (18 July 1635 - 3 March 1703). 

Definition 7.1: Hooke's Law 

In an elastic spring, the extension varies linearly with the force applied. 

F = -kx 

where F is the restoring force in newtons (N), k is the spring constant in N ■ m~^ and x is the 
displacement of the spring from its equilibrium length in metres (m) . 



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171 



172 CHAPTER 7. MECHANICAL PROPERTIES OF MATTER 



Image notjinished 



Figure 7.1: Hooke's Law - the relationship between a spring's restoring force and its displacement from 
equilibrium length. 



7.1.2.1.1 Experiment : Hooke's Law 

Aim: 

Verify Hooke's Law. 
Apparatus: 

• weights 

• spring 

• ruler 

Method: 

1. Set up a spring vertically in such a way that you are able to hang weights from it. 

2. Measure the equilibrium length, xq, of the spring (i.e. the length of the spring when nothing is attached 
to it). 

3. Measure the extension of the spring for a range of different weights. Note: the extension is the difference 
between the spring's equilibrium length and the new length when a weight is attached to it, x — xq. 

4. Draw a table of force (weight) in newtons and corresponding extension. 

5. Draw a graph of force versus extension for your experiment. 

Conclusions: 

1. What do you observe about the relationship between the applied force and the extension? 

2. Determine the gradient (slope) of the graph. 

3. Now calculate the spring constant for your spring. 

Phet simulation for Hooke's Law 

This media object is a Flash object. Please view or download it at 
<mass-spring-lab.swf> 

Figure 7.2 



Khan academy video on springs and Hooke's law^ 

This media object is a Flash object. Please view or download it at 
<http://www.youtube.com/v/ZzwuHS91dbY&rel=0&hl=en_US&feature=player_embedded&version=3> 

Figure 7.3 



173 



Exercise 7.1: Hooke's Law^ I (Solution on p. 179.) 

A spring is extended by 7 cm by a force of 56 N. 
Calculate the spring constant for this spring. 

Exercise 7.2: Hooke's Law II (Solution on p. 179.) 

A spring of length 20cm stretches to 24cm when a load of 0,6N is applied to it. 

1. Calculate the spring constant for the spring. 

2. Determine the extension of the spring if a load of 0,5N is applied to it. 

Exercise 7.3: Hooke's Law III (Solution on p. 179.) 

A spring has a spring constant of —400 N.m~^. By how much will it stretch if a load of 50 N is 
applied to it? 



7.1.2.2 Deviation from Hooke's Law^ 

We know that if you have a small spring and you pull it apart too much it stops 'working'. It bends out of 
shape and loses its springiness. When this happens, Hooke's Law no longer applies, the spring's behaviour 
deviates from Hooke's Law. 

Depending on what type of material we are dealing with, the manner in which it deviates from Hooke's 
Law is different. We give classify materials by this deviation. The following graphs show the relationship 
between force and extension for different materials and they all deviate from Hooke's Law. Remember that 
a straight line show proportionality so as soon as the graph is no longer a straight line, Hooke's Law no 
longer applies. 

7.1.2.2.1 Brittle material 



Image notjtnished 

Figure 7.4: A hard, brittle substance 



This graph shows the relationship between force and extension for a brittle, but strong material. Note that 
there is very little extension for a large force but then the material suddenly fractures. Brittleness is the 
property of a material that makes it break easily without bending. 

Have you ever dropped something made of glass and seen it shatter? Glass does this because it is brittle. 

7.1.2.2.2 Plastic material 



Image notjtnished 

Figure 7.5: A plastic material's response to an applied force. 



174 CHAPTER 7. MECHANICAL PROPERTIES OF MATTER 

Here the graph shows the relationship between force and extension for a plastic material. The material 
extends under a small force but it does not fracture easily, and it does not return to its original length when 
the force is removed. 

7.1.2.2.3 Ductile material 



Image notjtnished 

Figure 7.6: A ductile substance. 



In this graph the relationship between force and extension is for a material that is ductile. The material 
shows plastic behaviour over a range of forces before the material finally fractures. Ductility is the ability of 
a material to be stretched into a new shape without breaking. Ductility is one of the characteristic properties 
of metals. 

A good example of this is aluminium, many things are made of aluminium. Aluminium is used for making 
everything from cooldrink cans to aeroplane parts and even engine blocks for cars. Think about squashing 
and bending a cooldrink can. 

Brittleness is the opposite of ductility. 

When a material reaches a point where Hooke's Law is no longer valid, we say it has reached its limit of 
proportionality. After this point, the material will not return to its original shape after the force has been 
removed. We say it has reached its elastic limit. 

Definition 7.2: Elastic limit 

The elastic limit is the point beyond which permanent deformation takes place. 

Definition 7.3: Limit of proportionality 

The limit of proportionality is the point beyond which Hooke's Law is no longer obeyed. 

7.1.2.2.3.1 Hooke's Law and deformation of materials 

1. What causes deformation? 

2. Describe Hooke's Law in words and mathematically. 

3. List similarities and differences between ductile, brittle and plastic (polymeric) materials, with specific 
reference to their force-extension graphs. 

4. Describe what is meant by the elastic limit. 

5. Describe what is meant by the limit of proportionality. 

6. A spring of length 15 cm stretches to 27 cm when a load of 0,4 N is applied to it. 

a. Calculate the spring constant for the spring. 

b. Determine the extension of the spring if a load of 0,35 N is applied to it. 

7. A spring has a spring constant of —200 N.m~^. By how much will it stretch if a load of 25 N is applied 
to it? 

8. A spring of length 20 cm stretches to 24 cm when a load of 0,6 N is applied to it. 

a. Calculate the spring constant for the spring. 

b. Determine the extension of the spring if a load of 0,8 N is applied to it. 



175 

7.2 Failure and strength of materials' 
7.2,1 Elasticity, plasticity, fracture, creep 

7.2.1.1 Elasticity and plasticity 

Materials are classified as plastic or elastic depending on how they respond to an applied force. It is important 
to note that plastic substances are not necessarily a type of plastic (polymer) they only behave like plastic. 
Think of them as being like plastic which you will be familiar with. 

A rubber band is a material that has elasticity. It returns to its original shape after an applied force is 
removed, providing that the material is not stretched beyond its elastic limit. 

Plasticine is an example of a material that is plastic. If you fiatten a ball of plasticine, it will stay fiat. 
A plastic material does not return to its original shape after an applied force is removed. 

• Elastic materials return to their original shape. 

• Plastic materials deform easily and do not return to their original shape. 



7.2.1.2 Fracture, creep and fatigue 

Some materials are neither plastic nor elastic. These substances will break or fracture when a large enough 
force is applied to them. The brittle glass we mentioned earlier is an example. 

Creep occurs when a material deforms over a long period of time because of an applied force. An example 
of creep is the bending of a shelf over time when a heavy object is put on it. Creep may eventually lead to 
the material fracturing. The application of heat may lead to an increase in creep in a material. 

Fatigue is similar to creep. The difference between the two is that fatigue results from the force being 
applied and then removed repeatedly over a period of time. With metals this results in failure because of 
metal fatigue. 

• Fracture is an abrupt breaking of the material. 

• Creep is a slow deformation process due to a continuous force over a long time. 

• Fatigue is weakening of the material due to short forces acting many many times. 



7.2.1.2.1 Elasticity, plasticity, fracture and creep 

1. List the similarities and differences between elastic and plastic deformation. 

2. List the similarities and differences between creep and fracture as modes of failure in material. 



7.2,2 Failure and strength of materials 

7.2.2.1 The properties of matter 

The strength of a material is defined as the stress (the force per unit cross-sectional area) that it can 
withstand. Strength is measured in newtons per square metre {N ■ to~^). 

Stiffness is a measure of how fiexible a material is. In Science we measure the stiffness of a material by 
calculating its Young's Modulus. The Young's modulus is a ratio of how much it bends to the load applied 
to it. Stiffness is measure in newtons per metre {N ■ m~^). 

Hardness of a material can be measured by determining what force will cause a permanent deformation in 
the material. Hardness can also be measured using a scale like Mohs hardness scale. On this scale, diamond 
is the hardest at 10 and talc is the softest at 1. 



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176 CHAPTER 7. MECHANICAL PROPERTIES OF MATTER 

NOTE: Remembering that the Mohs scale is the hardness scale and that the softest substance is 
talc will often come in handy for general knowledge quizes. 

The toughness of a material is a measure of how it can resist breaking when it is stressed. It is scientifically 
defined as the amount of energy that a material can absorb before fracturing. 

A ductile material is a substance that can undergo large plastic deformation without fracturing. Many 
metals are very ductile and they can be drawn into wires, e.g. copper, silver, aluminium and gold. 

A malleable material is a substance that can easily undergo plastic deformation by hammering or rolling. 
Again, metals are malleable substances, e.g. copper can be hammered into sheets and aluminium can be 
rolled into aluminium foil. 

A brittle material fractures with very little or no plastic deformation. Glassware and ceramics are brittle. 

7.2.2.2 Structure and failure of materials 

Many substances fail because they have a weakness in their atomic structure. There are a number of 
problems that can cause these weaknesses in structure. These are vacancies, dislocations, grain boundaries 
and impurities. 

Vacancies occur when there are spaces in the structure of a crystalline solid. These vacancies cause 
weakness and such materials often fail at these places. Think about bricks in a wall, if you started removing 
bricks the wall would get weaker. 

Dislocations result in weakened bonding between layers of atoms in a crystal lattice and this creates a 
critical boundary. If sufficient force is applied along the boundary, it can break the weakened bonds, allowing 
the two sides of the crystal to slide against one another. The two pieces of the crystal keep their shape and 
structure. 

Impurities in a crystal structure can cause a weak region in the crystal lattice around the impurity. Like 
vacancies, the substance often fail from these places in the lattice. This you can think of as bricks in a wall 
which don't fit properly, they are the wrong kind of bricks (atoms) to make the structure strong. 

7.2.2.3 Controlling the properties of materials 

There are a number of processes that can be used to make materials less likely to fail. We shall look at a 
few methods in this section. 

7.2.2.4 Cold working 

Cold working is a process in which a metal is strengthened by repeatedly being reshaped. This is carried 
out at a temperature below the melting point of the metal. The repeated shaping of the metal results in 
dislocations which then prevent restrict the motion of dislocations in the metal. Cold working increases the 
strength of the metal but in so doing, the metal loses its ductility. We say the metal is work-hardened. 

7.2.2.5 Annealing 

Annealing is a process of heating and cooling a material to relax the crystal structure and reduce weakness 
due to impurities and structural flaws. During annealing, the material is heated to a high temperature 
that is below the material's melting point. At a sufficiently high temperature, atoms with weakened bonds 
can rearrange themselves into a stronger structure. Slowly cooling the material ensures that the atoms will 
remain in these stronger locations. Annealing is often used before cold working. 

7.2.2.6 Introduction of Impurities 

Most pure metals are relatively weak because dislocations can move easily within them. However, if impurities 
are added to a metal (e.g., carbon is added to an iron sample), they can disturb the regular structure of the 
metal and so prevent dislocations from spreading. This makes the metal stronger. 



177 

7.2.2.7 Alloying 

An alloy is a mixture of a metal with other substances. In other words, alloying involves adding impurities 
to a metal sample. The other substances can be metal or non-metal. An alloy often has properties that are 
very different to the properties of the substances from which it is made. The added substances strengthen 
the metal by preventing dislocations from spreading. Ordinary steel is an alloy of iron and carbon. The 
carbon impurities trap dislocations. There are many types of steel that also include other metals with iron 
and carbon. Brass is an alloy of copper and Zinc. Bronze is an alloy of copper and tin. Gold and silver that 
is used in coins or jewellery are also alloyed. 

7.2.2.8 Tempering 

Tempering is a process in which a metal is melted then quickly cooled. The rapid cooling is called quenching. 
Usually tempering is done a number of times before a metal has the correct properties that are needed for 
a particular application. 

7.2.2.9 Sintering 

Sintering is used for making ceramic objects among other things. In this process the substance is heated so 
that its particles stick together. It is used with substances that have a very high melting point. The resulting 
product is often very pure and it is formed in the process into the shape that is wanted. Unfortunately, 
sintered products are brittle. 

7.2.2.10 Steps of Roman Swordsmithing 

• Purifying the iron ore. 

• Heating the iron blocks in a furnace with charcoal. 

• Hammering and getting into the needed shape. The smith used a hammer to pound the metal into 
blade shape. He usually used tongs to hold the iron block in place. 

• Reheating. When the blade cooled, the smith reheated it to keep it workable. While reheated and 
hammered repeatedly. 

• Quenching which involved the process of white heating and cooling in water. Quenching made the 
blade harder and stronger. At the same time it made the blade quite brittle, which was a considerable 
problem for the sword smiths. 

• Tempering was then done to avoid brittleness the blade was tempered. In another words it was reheated 
a final time to a very specific temperature. How the Romans do balanced the temperature? The smith 
was guided only by the blade's color and his own experience. 



7.2.2.10.1 Failure and strength of materials 

1. List the similarities and differences between the brittle and ductile modes of failure. 

2. What is meant by the following terms: 

a. vacancies 

b. dislocations 

c. impurities 

d. grain boundaries 

3. What four terms can be used to describe a material's mechanical properties? 

4. What is meant by the following: 

a. cold working 

b. annealing 

c. tempering 



178 CHAPTER 7. MECHANICAL PROPERTIES OF MATTER 

d. introduction of impurities 

e. alloying 

f. sintering 



7.2,3 Summary 

1. Hooke's Law gives the relationship between the extension of a spring and the force applied to it. The 
law says they are proportional. 

2. Materials can be classified as plastic or elastic depending on how they respond to an applied force. 

3. Materials can fracture or undergo creep or fatigue when forces are applied to them. 

4. Materials have the following mechanical properties to a greater or lesser degree: strength, hardness, 
ductility, malleability, brittleness, stiffness. 

5. Materials can be weakened by have the following problems in their crystal lattice: vacancies, disloca- 
tions, impurities, difference in grain size. 

6. Materials can have their mechanical properties improved by one or more of the following processes: 
cold working, annealing, adding impurities, tempering, sintering. 



7.2,4 End of chapter exercise 

1. State Hooke's Law in words. 

2. What do we mean by the following terms with respect to Hooke's Law? 

a. elastic limit 

b. limit of proportionality 

3. A spring is extended by 18 cm by a force of 90 N. Calculate the spring constant for this spring. 

4. A spring of length 8 cm stretches to 14 cm when a load of 0,8 N is applied to it. 

a. Calculate the spring constant for the spring. 

b. Determine the extension of the spring if a load of 0,7 N is applied to it. 

5. A spring has a spring constant of —150 N.m~^. By how much will it stretch if a load of 80 N is applied 
to it? 

6. What do we mean by the following terms when speaking about properties of materials? 

a. hardness 

b. toughness 

c. ductility 

d. malleability 

e. stiffness 

f. strength 

7. What is Young's modulus? 

8. In what different ways can we improve the material properties of substances? 

9. What is a metal alloy? 

10. What do we call an alloy of: 

a. iron and carbon 

b. copper and zinc 

c. copper and tin 

11. Do some research on what added substances can do to the properties of steel. Present your findings in 
a suitable table. 



179 



Solutions to Exercises in Chapter 7 

Solution to Exercise 7.1 (p. 173) 

Step 1. 



F = 


-kx 


56 = 


-k X 0,07 


; = 


-56 
0,07 


= - 


-800iV-m- 



Solution to Exercise 7.2 (p. 173) 

Step 1. We know: 

i^ = 0,6 N 

The equilibrium spring length is 20 cm 

The expanded spring length is 24 cm 
Step 2. First we need to calculate the displacement of the spring from its equilibrium length: 













X = 24cm - 20cm 
















= 4cm 
















= 0, 04m 






use 


Hooke's 


Law to 


find the 


spring constant: 
















F = 


kx 














0,6 = -k- 


0, 


04 










k = 


-IbN.m-^ 







Step 3. F = 0,5 N 

We know from the first part of the question that 

k = -15 N.m'^'^ 

So, using Hooke's Law: 



F = -kx 

- = -f 

— 0,5 

~ -15 

= 0,033m 

= 3, 3cm 



Solution to Exercise 7.3 (p. 173) 






Step 1. 








F = 


-kx 




50 = 


-(-400) a; 




X = 


50 
400 




= 


0,125m 




= 


12,5cm 



(7.1) 
(7.2) 



(7.3) 



(7.4) 



(7.5) 



(7.6) 



(7.7) 



180 CHAPTER 7. MECHANICAL PROPERTIES OF MATTER 



Chapter 8 

Work, energy and power 



8.1 Work^ 

8.1.1 Introduction 

Imagine a vendor carrying a basket of vegetables on her head. Is she doing any work? One would definitely 
say yes! However, in Physics she is not doing any work! Again, imagine a boy pushing against a wall? Is 
he doing any work? We can see that his muscles are contracting and expanding. He may even be sweating. 
But in Physics, he is not doing any work! 

If the vendor is carrying a very heavy load for a long distance, we would say she has lot of energy. By 
this, we mean that she has a lot of stamina. If a car can travel very fast, we describe the car as powerful. 
So, there is a link between power and speed. However, power means something different in Physics. This 
chapter describes the links between work, energy and power and what these mean in Physics. 

You will learn that work and energy are closely related. You shall see that the energy of an object is its 
capacity to do work and doing work is the process of transferring energy from one object or form to another. 
In other words, 

• an object with lots of energy can do lots of work. 

• when work is done, energy is lost by the object doing work and gained by the object on which the work 
is done. 

Lifting objects or throwing them requires that you do work on them. Even making electricity fiow requires 
that something do work. Something must have energy and transfer it through doing work to make things 
happen. 

8.1.2 Work 

Definition 8.1: Work 

When a force exerted on an object causes it to move, work is done on the object (except if the 
force and displacement are at right angles to each other). 

This means that in order for work to be done, an object must be moved a distance rf by a force F, such 
that there is some non-zero component of the force in the direction of the displacement. Work is calculated 
as: 

W = F ■ Axcose. (8.1) 

where F is the applied force. Ax is the displacement of the object and 9 is the angle between the applied 
force and the direction of motion. 



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181 



182 CHAPTER 8. WORK, ENERGY AND POWER 



Image notjinished 

Figure 8.1: The force F causes the object to be displaced by Ax at angle 9. 



It is very important to note that for work to be done there must be a component of the applied force in 
the direction of motion. Forces perpendicular to the direction of motion do no work. 
For example work is done on the object in Figure 8.2, 



Image notjinished 



Figure 8.2: (a) The force F causes the object to be displaced by Ax in the same direction as the force. 
9 — 0° and cos9 — 1. Work is done in this situation, (b) A force F is applied to the object. The object 
is displaced by Ay at right angles to the force. 9 — 90° and cos9 — 0. Work is not done in this situation. 



8.1.2.1 Investigation : Is work done? 

Decide whether on not work is done in the following situations. Remember that for work to be done a force 
must be applied in the direction of motion and there must be a displacement. Give reasons for your answer. 

1. Max pushes against a wall and becomes tired. 

2. A book falls off a table and free falls to the ground. 

3. A rocket accelerates through space. 

4. A waiter holds a tray full of meals above his head with one arm and carries it straight across the room 
at constant speed. (Careful! This is a tricky question.) 

TIP: The Meaning of 9: The angle 6 is the angle between the force vector and the displacement 
vector. In the following situations, 9 = 0°. 



Image notjinished 

Figure 8.3 



As with all physical quantities, work must have units. Following from the definition, work is measured in 
N-m. The name given to this combination of S.I. units is the joule (symbol J). 

Definition 8.2: Joule 

1 joule is the work done when an object is moved 1 m under the application of a force of 1 N in 
the direction of motion. 



183 

The work done by an object can be positive or negative. Since force (Fy) and displacement (s) are both 
vectors, the result of the above equation depends on their directions: 

• If F|| acts in the same direction as the motion then positive work is being done. In this case the object 
on which the force is applied gains energy. 

• If the direction of motion and Fy are opposite, then negative work is being done. This means that 
energy is transferred in the opposite direction. For example, if you try to push a car uphill by applying 
a force up the slope and instead the car rolls down the hill you are doing negative work on the car. 
Alternatively, the car is doing positive work on you! 

TIP: The everyday use of the word "work" differs from the physics use. In physics, only the 
component of the applied force that is parallel to the motion does work on an object. So, for 
example, a person holding up a heavy book does no work on the book. 

Exercise 8.1: Calculating Work Done I (Solution on p. 195.) 

If you push a box 20 m forward by applying a force of 15 N in the forward direction, what is the 
work you have done on the box? 

Exercise 8.2: Calculating Work Done II (Solution on p. 195.) 

What is the work done by you on a car, if you try to push the car up a hill by applying a force of 
40 N directed up the slope, but it slides downhill 30 cm? 

What happens when the applied force and the motion are not parallel? If there is an angle between the 
direction of motion and the applied force then to determine the work done we have to calculate the component 
of the applied force parallel to the direction of motion. Note that this means a force perpendicular to the 
direction of motion can do no work. 

Exercise 8.3: Calculating Work Done III (Solution on p. 195.) 

Calculate the work done on a box, if it is pulled 5 m along the ground by applying a force of 
_F=10 N at an angle of 60° to the horizontal. 



Image notjinished 



Figure 8.4 



8.1.2.2 Work 

1. A 10 N force is applied to push a block across a friction free surface for a displacement of 5.0 m to the 
right. The block has a weight of 20 N. Determine the work done by the following forces: normal force, 
weight, applied force. 



Image notjinished 

Figure 8.5 



184 CHAPTER 8. WORK, ENERGY AND POWER 

2. A 10 N frictional force slows a moving block to a stop after a displacement of 5.0 m to the right. The 
block has a weight of 20 N. Determine the work done by the following forces: normal force, weight, 
frictional force. 

Image notjinished 

Figure 8.6 



3. A 10 N force is applied to push a block across a frictional surface at constant speed for a displacement 
of 5.0 m to the right. The block has a weight of 20 N and the frictional force is 10 N. Determine the 
work done by the following forces: normal force, weight, frictional force. 



Image notjinished 



Figure 8.7 



4. A 20 N object is sliding at constant speed across a friction free surface for a displacement of 5 m to 
the right. Determine if there is any work done. 



Image notjinished 

Figure 8.8 



5. A 20 N object is pulled upward at constant speed by a 20 N force for a vertical displacement of 5 m. 
Determine if there is any work done. 



Im.age notjinished 



Figure 8.9 



6. Before beginning its descent, a roller coaster is always pulled up the first hill to a high initial height. 
Work is done on the roller coaster to achieve this initial height. A coaster designer is considering three 
different incline angles of the hill at which to drag the 2 000 kg car train to the top of the 60 m high 
hill. In each case, the force applied to the car will be applied parallel to the hill. Her critical question 
is: which angle would require the least work? Analyze the data, determine the work done in each case, 
and answer this critical question. 



185 



Angle of Incline 


Applied Force 


Distance 


Work 


35° 


1.1 X W^N 


100 m 




45° 


1.3 X W^N 


90 m 




55° 


1.5 X lO'^iV 


80 m 





Table 8.1 

7. Big Bertha carries a 150 N suitcase up four flights of stairs (a total height of 12 m) and then pushes 
it with a horizontal force of 60 N at a constant speed of 0.25 m-s~^ for a horizontal distance of 50 m 
on a frictionless surface. How much work does Big Bertha do on the suitcase during this entire trip? 

8. A mother pushes down on a pram with a force of 50 N at an angle of 30°. The pram is moving on a 
frictionless surface. If the mother pushes the pram for a horizontal distance of 30 m, how much does 
she do on the pram? 



Image notjtnished 



Figure 8.10 



9. How much work is done by an applied force to raise a 2 000 N lift 5 floors vertically at a constant 
speed? Each floor is 5 m high. 
10. A student with a mass of 60 kg runs up three flights of stairs in 15 s, covering a vertical distance of 
10 m. Determine the amount of work done by the student to elevate her body to this height. Assume 
that her speed is constant. 



8.2 Energy' 
8.2,1 Energy 

8.2.1.1 External and Internal Forces 

In Grade 10, you saw that mechanical energy was conserved in the absence of external forces. It is important 
to know whether a force is an internal force or an external force in the system, because this is related to 
whether the force can change an object's total mechanical energy when it does work on an object. 

When an external force (for example friction, air resistance, applied force) does work on an object, the 
total mechanical energy (KE + PE) of that object changes. If positive work is done, then the object will 
gain energy. If negative work is done, then the object will lose energy. The gain or loss in energy can be in 
the form of potential energy, kinetic energy, or both. However, the work which is done is equal to the change 
in mechanical energy of the object. 



8.2.1.1.1 Investigations : External Forces 

We can investigate the effect of external forces on an object's total mechanical energy by rolling a ball along 
the floor from point A to point B. 



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186 CHAPTER 8. WORK, ENERGY AND POWER 

Image notjinished 

Figure 8.11 



Find a nice smooth surface (e.g. a highly polished floor), mark off two positions, A and B, and roll the 
ball between them. 

The total mechanical energy of the ball, at each point, is the sum of its kinetic energy (KE) and gravi- 
tational potential energy (PE): 



StotaM = KEa + PEa 

(8.2) 



\Tnv\ + mghA 
\mv\ + mg (0) 



2'"'' A' 



'A 



Etota\,B = KEb + PEb 

= -^mv^g + mghB 



.3) 



^mv'^ + mg (0) 



1 2 



VA 


^ 


VB 


\mv\ 


= 


\mv\ 


'total, A 


= 


Etotal.B 



In the absence of friction and other external forces, the ball should slide along the floor and its speed 
should be the same at positions A and B. Since there are no external forces acting on the ball, its total 
mechanical energy at points A and B are equal. 



?.4) 



Now, let's investigate what happens when there is friction (an external force) acting on the ball. 

Roll the ball along a rough surface or a carpeted floor. What happens to the speed of the ball at point 
A compared to point B? 

If the surface you are rolling the ball along is very rough and provides a large external frictional force, 
then the ball should be moving much slower at point B than at point A. 

Let's now compare the total mechanical energy of the ball at points A and B: 

StotaM = KEa + PEa 
= ^mv'i + mghA 
— mvA + fng (0) 



2 



^mvl 



Etota\,B = KEb + PEb 

= ^mvg + mghB 



.6) 



jTTiv'^ + mg (0) 



1 2 



187 



However, in this case, va 7^ vb and therefore Etoted.A / £'totai,B- Since 

"^ > "^ (8.7) 

£'totaKA > £'total,B 

Therefore, the ball has lost mechanical energy as it moves across the carpet. However, although the ball 
has lost mechanical energy, energy in the larger system has still been conserved. In this case, the missing 
energy is the work done by the carpet through applying a frictional force on the ball. In this case the carpet 
is doing negative work on the ball. 

When an internal force does work on an object by an (for example, gravitational and spring forces), the 
total mechanical energy (KE + PE) of that object remains constant but the object's energy can change form. 
For example, as an object falls in a gravitational field from a high elevation to a lower elevation, some of 
the object's potential energy is changed into kinetic energy. However, the sum of the kinetic and potential 
energies remain constant. When the only forces doing work are internal forces, energy changes forms - from 
kinetic to potential (or vice versa); yet the total amount of mechanical energy is conserved. 

8.2.1.2 Capacity to do Work 

Energy is the capacity to do work. When positive work is done on an object, the system doing the work loses 
energy. In fact, the energy lost by a system is exactly equal to the work done by the system. An 

object with larger potential energy has a greater capacity to do work. 

Exercise 8.4: Work Done on a System (Solution on p. 196.) 

Show that a hammer of mass 2 kg does more work when dropped from a height of 10 m than when 
dropped from a height of 5 m. Confirm that the hammer has a greater potential energy at 10 m 
than at 5 m. 

This leads us to the work-energy theorem. 

Definition 8.3: Work-Energy Theorem 

The work-energy theorem states that the work done on an object is equal to the change in its 
kinetic energy: 

W = AKE = KEf - KEi (8.8) 

The work-energy theorem is another example of the conservation of energy which you saw in Grade 10. 

Exercise 8.5: Work-Energy Theorem (Solution on p. 197.) 

A brick of mass 1 kg is dropped from a height of 10 m. Calculate the work done on the brick at 
the point it hits the ground assuming that there is no air resistance? 

Exercise 8.6: Work-Energy Theorem 2 (Solution on p. 197.) 

The driver of a 1 000 kg car traveling at a speed of 16,7 m • s~^ applies the car's brakes when 
he sees a red robot. The car's brakes provide a frictional force of 8000 N. Determine the stopping 
distance of the car. 



TIP: A force only does work on an object for the time that it is in contact with the object. For 
example, a person pushing a trolley does work on the trolley, but the road does no work on the 
tyres of a car if they turn without slipping (the force is not applied over any distance because a 
different piece of tyre touches the road every instant. 

TIP: Energy is conserved! 



188 CHAPTER 8. WORK, ENERGY AND POWER 

TIP: In the absence of friction, the work done on an object by a system is equal to the energy 
gained by the object. 

Work Done = Energy Transferred (8-9) 

In the presence of friction, only some of the energy lost by the system is transferred to useful 
energy. The rest is lost to friction. 

Total work done = Useful work done + Work done against friction (8.10) 

In the example of a falling mass the potential energy is known as gravitational potential energy as it is 
the gravitational force exerted by the earth which causes the mass to accelerate towards the ground. The 
gravitational field of the earth is what does the work in this case. 

Another example is a rubber-band. In order to stretch a rubber-band we have to do work on it. This 
means we transfer energy to the rubber-band and it gains potential energy. This potential energy is called 
elastic potential energy. Once released, the rubber-band begins to move and elastic potential energy is 
transferred into kinetic energy. 

8.2.1.2.1 Other forms of Potential Energy 

1. elastic potential energy - potential energy is stored in a compressed or extended spring or rubber band. 
This potential energy is calculated by: 

-kx"^ (8.11) 

where fc is a constant that is a measure of the stiffness of the spring or rubber band and x is the 
extension of the spring or rubber band. 

2. Chemical potential energy is related to the making and breaking of chemical bonds. For example, a 
battery converts chemical energy into electrical energy. 

3. The electrical potential energy of an electrically charged object is defined as the work that must be done 
to move it from an infinite distance away to its present location, in the absence of any non-electrical 
forces on the object. This energy is non-zero if there is another electrically charged object nearby 
otherwise it is given by: 

k^-^ (8.12) 

d 

where k is Coulomb's constant. For example, an electric motor lifting an elevator converts electrical 
energy into gravitational potential energy. 

4. Nuclear energy is the energy released when the nucleus of an atom is split or fused. A nuclear reactor 
converts nuclear energy into heat. 

Some of these forms of energy will be studied in later chapters. 

8.2.1.2.2 Investigation : Energy Resources 

Energy can be taken from almost anywhere. Power plants use many different types of energy sources, 
including oil, coal, nuclear, biomass (organic gases), wind, solar, geothermal (the heat from the earth's rocks 
is very hot underground and is used to turn water to steam), tidal and hydroelectric (waterfalls). Most power 
stations work by using steam to turn turbines which then drive generators and create an electric current. 

Most of these sources are dependant upon the sun's energy, because without it we would not have weather 
for wind and tides. The sun is also responsible for growing plants which decompose into fossil fuels like oil 
and coal. All these sources can be put under 2 headings, renewable and non-renewable. Renewable sources 
are sources which will not run out, like solar energy and wind power. Non-renewable sources are ones which 
will run out eventually, like oil and coal. 



189 

It is important that we learn to appreciate conservation in situations like this. The planet has a number 
of linked systems and if we don't appreciate the long-term consequences of our actions we run the risk of 
doing damage now that we will only suffer from in many years time. 

Investigate two types of renewable and two types of non-renewable energy resources, listing advantages 
and disadvantages of each type. Write up the results as a short report. 

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Figure 8.12 



8.2.1.2.3 Energy 

1. Fill in the table with the missing information using the positions of the ball in the diagram below 
combined with the work-energy theorem. 



Image notjinished 



Figure 8.13 



position 


KE 


PE 


V 


A 




50 J 




B 




30 J 




C 








D 




10 J 




E 








F 








G 









Table 8.2 



2. A falling ball hits the ground at 10 m-s~^ in a vacuum. Would the speed of the ball be increased or 
decreased if air resistance were taken into account. Discuss using the work-energy theorem. 

3. A pendulum with mass 300g is attached to the ceiling. It is pulled up to point A which is a height h 
= 30 cm from the equilibrium position. 



190 CHAPTER 8. WORK, ENERGY AND POWER 

Image notjinished 

Figure 8.14 



Calculate the speed of the pendulum when it reaches point B (the equilibrium point). Assume that 
there are no external forces acting on the pendulum. 



8.3 Power' 

8.3.1 Power 

Now that we understand the relationship between work and energy, we are ready to look at a quantity that 
defines how long it takes for a certain amount of work to be done. For example, a mother pushing a trolley 
full of groceries can take 30 s or 60 s to push the trolley down an aisle. She does the same amount of work, 
but takes a different length of time. We use the idea of power to describe the rate at which work is done. 

Definition 8.4: Power 

Power is defined as the rate at which work is done or the rate at which energy is expended. The 
mathematical definition for power is: 

P = F -v (8.13) 

(8.13) is easily derived from the definition of work. We know that: 

W = F-d. (8.14) 

However, power is defined as the rate at which work is done. Therefore, 



P.- 



This can be written as: 



P 



AW 

At 
A(F-d) 

^* (8.16) 

pAd ^ ^ 

^Ai 



The unit of power is watt (symbol W) . 
8.3.1.1 Investigation : Watt 



Show that the W is equivalent to J • s ^. 

NOTE: The unit watt is named after Scottish inventor and engineer James Watt (19 January 1736 
- 19 August 1819) whose improvements to the steam engine were fundamental to the Industrial 
Revolution. A key feature of it was that it brought the engine out of the remote coal fields into 
factories. 



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191 

8.3.1.2 Research Project : James Watt 

Write a short report 5 pages on the Hfe of James Watt describing his many other inventions. 

NOTE: Historically, the horsepower (symbol hp) was the unit used to describe the power delivered 
by a machine. One horsepower is equivalent to approximately 750 W. The horsepower is sometimes 
used in the motor industry to describe the power output of an engine. Incidentally, the horsepower 
was derived by James Watt to give an indication of the power of his steam engine in terms of the 
power of a horse, which was what most people used to for example, turn a mill wheel. 

Exercise 8.7: Power Calculation 1 (Solution on p. 198.) 

Calculate the power required for a force of 10 N applied to move a 10 kg box at a speed of 1 ms 
over a frictionless surface. 

Machines are designed and built to do work on objects. All machines usually have a power rating. The 
power rating indicates the rate at which that machine can do work upon other objects. 

A car engine is an example of a machine which is given a power rating. The power rating relates to how 
rapidly the car can accelerate. Suppose that a 50 kW engine could accelerate the car from km • hr~ to 
60km • hr~^ in 16 s. Then a car with four times the power rating (i.e. 200 kW) could do the same amount 
of work in a quarter of the time. That is, a 200 kW engine could accelerate the same car from km • hr~^ 
to 60km • hr^^ in 4 s. 

Exercise 8.8: Power Calculation 2 (Solution on p. 199.) 

A forklift lifts a crate of mass 100 kg at a constant velocity to a height of 8 m over a time of 4 s. 
The forklift then holds the crate in place for 20 s. Calculate how much power the forklift exerts in 
lifting the crate? How much power does the forklift exert in holding the crate in place? 



8.3.1.3 Experiment : Simple measurements of human pow^er 

You can perform various physical activities, for example lifting measured weights or climbing a flight of stairs 
to estimate your output power, using a stop watch. Note: the human body is not very efficient in these 
activities, so your actual power will be much greater than estimated here. 

8.3.1.4 Power 

lEB 2005/11 HG Which of the following is equivalent to the SI unit of power: 

a. V-A 

b. V-A-i 

c. kg ■ m ■ s^^ 

d. kg • m -s"^ 

1. Two students. Bill and Bob, are in the weight lifting room of their local gum. Bill lifts the 50 kg barbell 
over his head 10 times in one minute while Bob lifts the 50 kg barbell over his head 10 times in 10 
seconds. Who does the most work? Who delivers the most power? Explain your answers. 

2. Jack and Jill ran up the hill. Jack is twice as massive as Jill; yet Jill ascended the same distance in 
half the time. Who did the most work? Who delivered the most power? Explain your answers. 

3. Alex (mass 60 kg) is training for the Comrades Marathon. Part of Alex's training schedule involves 
push-ups. Alex does his push-ups by applying a force to elevate his center-of-mass by 20 cm. Determine 
the number of push-ups that Alex must do in order to do 10 J of work. If Alex does all this work in 
60 s, then determine Alex's power. 

4. When doing a chin-up, a physics student lifts her 40 kg body a distance of 0.25 m in 2 s. What is the 
power delivered by the student's biceps? 



192 CHAPTER 8. WORK, ENERGY AND POWER 

5. The unit of power that is used on a monthly electricity account is kilowatt-hours (symbol kWh). This 
is a unit of energy delivered by the flow of 1 kW of electricity for 1 hour. Show how many joules of 
energy you get when you buy 1 kWh of electricity. 

6. An escalator is used to move 20 passengers every minute from the first floor of a shopping mall to the 
second. The second floor is located 5-meters above the first floor. The average passenger's mass is 
70 kg. Determine the power requirement of the escalator in order to move this number of passengers 
in this amount of time. 

7. Calculate the power required for an electric motor to pump 20kg of water up to ground level from a 
borehole of depth 10m in half a minute. 

These two videos provide a summary of some of the concepts covered in this chapter. 

Khan academy video on ^vork and energy - 1 

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Figure 8.15 



Khan academy video on work and energy - 2 

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Figure 8.16 



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Figure 8.17 



193 



8.3.2 Important Equations and Quantities 



Units 


Quantity 


Symbol 


Unit 


S.I. Units 


Direction 


velocity 


V 


— 


s 


or m.s^^ 




momentum 


P 


— 


kg.m 
s 


or kg.m.s^^ 




energy 


E 


J 


kg.rn^ 


or kg.TTi^s^^ 


— 


s^ 


Work 


W 


J 


N.m 


or kg.m^.s^^ 


— 


Kinetic Energy 


Ek 


J 


N.m 


or kg.m^.s^^ 


— 


Potential Energy 


Ep 


J 


N.m 


or kg.m'^.s^^ 


— 


Mechanical Energy 


U 


J 


N.m 


or kg.m'^.s^^ 


— 


Power 


P 


W 


N.m.s-^ 


or kg.m'^.s^^ 


— 



Table 8.3: Commonly used units 



Momentum: 



Kinetic energy: 



P= m V 



1 ^2 
Ek= -mv 



.17) 



.18) 



Principle of Conservation of Energy:: Energy is never created nor destroyed, but is merely transformed 
from one form to another. 

Conservation of Mechanical Energy:: In the absence of friction, the total mechanical energy of an ob- 
ject is conserved. 

When a force moves in the direction along which it acts, work is done. 
Work is the process of converting energy. 
Energy is the ability to do work. 

8.3.3 End of Chapter Exercises 

1. The force vs. displacement graph shows the amount of force applied to an object by three different 
people. Abdul applies force to the object for the first 4 m of its displacement, Beth applies force 
from the 4 m point to the 6 m point, and Charles applies force from the 6 m point to the 8 m point. 
Calculate the work done by each person on the object? Which of the three does the most work on the 
object? 



Image notjinished 



Figure 8.18 



194 CHAPTER 8. WORK, ENERGY AND POWER 

2. How much work does a person do in pushing a shopping trolley with a force of 200 N over a distance 
of 80 m in the direction of the force? 

3. How much work does the force of gravity do in pulling a 20 kg box down a 45° frictionless inclined 
plane of length 18 m? 

lEB 2001/11 HGl Of which one of the following quantities is kg.m^.s""^ the base S.I. unit? 

a. Energy 

b. Force 

c. Power 

d. Momentum 

lEB 2003/11 HGl A motor is used to raise a mass m through a vertical height h in time t. What is the power of the 
motor while doing this? 

a. mght 

t 

(. mgt 

h 

d. -^ 

mg 

lEB 2002/11 HGl An electric motor lifts a load of mass M vertically through a height h at a constant speed v. Which of 
the following expressions can be used to correctly calculate the power transferred by the motor to the 
load while it is lifted at a constant speed? 

a. Mgh 

b. Mgh + iMv2 

c. Mgv 

d. Mgv+ iMx! 

lEB 2001/11 HGl An escalator is a moving staircase that is powered by an electric motor. People are lifted up the 
escalator at a constant speed of v through a vertical height h. What is the energy gained by a person 
of mass m standing on the escalator when he is lifted from the bottom to the top? 



Image notjinished 

Figure 8.19 



a. mgh 

b. mgh sin6 

d. 



,2 



lEB 2003/11 HGl In which of the following situations is there no work done on the object? 

a. An apple falls to the ground. 

b. A brick is lifted from the ground to the top of a building. 

c. A car slows down to a stop. 

d. A box moves at constant velocity across a frictionless horizontal surface. 



195 

Solutions to Exercises in Chapter 8 

Solution to Exercise 8.1 (p. 183) 

Step 1. • The force applied is F=15 N. 

• The distance moved is s=20 m. 

• The appHed force and distance moved are in the same direction. Therefore, f|j=15 N. 
These quantities are all in the correct units, so no unit conversions are required. 

Step 2. • We are asked to find the work done on the box. We know from the definition that work done is 

W = Fiis 
Step 3. 

W = F||s 

= (15 N){20 m) (8.19) 

300 J 

Remember that the answer must be positive as the applied force and the motion are in the same 
direction (forwards). In this case, you (the pusher) lose energy, while the box gains energy. 



Solution to Exercise 8.2 (p. 183) 

Step 1. • The force applied is F=40 N 



• 



The distance moved is s=30 cm. This is expressed in the wrong units so we must convert to the 
proper S.I. units (meters): 

100 cm = Im 

1 cm = j^ m 

.-.30x1 cm = ^Oxj^m (8.20) 

100 '"■ 
= 0,3m 

• The applied force and distance moved are in opposite directions. Therefore, if we take s=0.3 m, 
then F|[=-40 N. 

Step 2. • We are asked to find the work done on the car by you. We know that work done is W = F||S 
Step 3. Again we have the applied force and the distance moved so we can proceed with calculating the work 
done: 

W = F,[S 

= (-40 TV) (0.3m) (8.21) 

= -12J 

Note that the answer must be negative as the applied force and the motion are in opposite directions. 
In this case the car does work on the person trying to push. 

Solution to Exercise 8.3 (p. 183) 

Step 1. • The force applied is i^=10 N 

• The distance moved is s=5 m along the ground 

• The angle between the applied force and the motion is 60° 

These quantities are in the correct units so we do not need to perform any unit conversions. 
Step 2. • We are asked to find the work done on the box. 



196 CHAPTER 8. WORK, ENERGY AND POWER 

Step 3. Since the force and the motion are not in the same direction, we must first calculate the component of 
the force in the direction of the motion. 

Image notjtnished 

Figure 8.20 



From the force diagram we see that the component of the applied force parallel to the ground is 

(8.22) 

















F|| = 


F ■ cos (e 


)0°) 
















= 


10 A^ • cos 


(60° 
















= 


5N 




Step 


4. 


Now 


we 


can 


calculate 


the work done 


on the box 
W = 


(5 TV) (5 
25 J 


m) 



(8.23) 

Note that the answer is positive as the component of the force Fy is in the same direction as the 
motion. 

Solution to Exercise 8.4 (p. 187) 

Step 1. We are given: 

• the mass of the hammer, m =2 kg 

• height 1, hi=10 m 

• height 2, /i2=5 m 

We are required to show that the hammer does more work when dropped from hi than from /12. We 
are also required to confirm that the hammer has a greater potential energy at 10 m than at 5 m. 
Step 2. a. Calculate the work done by the hammer, Wi, when dropped from hi using: 

Wi = Fg-hi. (8.24) 

b. Calculate the work done by the hammer, W2, when dropped from /12 using: 

W2 = Fg- /i2. (8.25) 

c. Compare Wi and W2 

d. Calculate potential energy at hi and /12 and compare using: 

PE = m-g-h. (8.26) 

Step 3. 

Wi = Fg-hi 

= m ■ g ■ hi , ^ 

(8.27) 
= {2kg) {9.8 m-s-^) {10 m) 

196 J 



197 



Step 4. 

W2 = Fg-h2 

= m ■ g ■ h2 



(8.28) 
(2 kg) (9.8m-s-2) (5 m) 

98 J 



Step 5. We have T^i=196 J and W^2=98 J. Wi > Wi as required. 
Step 6. From (8.26), we see that: 

PEi = m ■ g ■ h 

= Fg-h (8.29) 

W 

This means that the potential energy is equal to the work done. Therefore, PEi > PE2, because 
Wi > W2. 

Solution to Exercise 8.5 (p. 187) 

Step 1. We are given: 

• mass of the brick: m=l kg 

• initial height of the brick: hi=10 m 

• final height of the brick: hf=0 m 

We are required to determine the work done on the brick as it hits the ground. 
Step 2. The brick is falling freely, so energy is conserved. We know that the work done is equal to the difference 

in kinetic energy. The brick has no kinetic energy at the moment it is dropped, because it is stationary. 

When the brick hits the ground, all the brick's potential energy is converted to kinetic energy. 
Step 3. 

PE = m ■ g ■ h 

= (1kg) (9,8m -8-2) (lOm) (8.30) 

98 J 

Step 4. The brick had 98 J of potential energy when it was released and J of kinetic energy. When the brick 
hit the ground, it had J of potential energy and 98 J of kinetic energy. Therefore KEi=0 J and 
KEf=98 J. 



From the work-energy theorem: 



W = ARE 

= KEf - KE, 



(8.31) 
98 J - J 

98 J 



Step 5. 98 J of work was done on the brick. 

Solution to Exercise 8.6 (p. 187) 
Step 1. We are given: 

• mass of the car: m=l 000 kg 

• speed of the car: w=16,7 m ■ s~^ 

• frictional force of brakes: F=8 000 N 



198 CHAPTER 8. WORK, ENERGY AND POWER 

We are required to determine the stopping distance of the car. 
Step 2. We apply the work-energy theorem. We know that all the car's kinetic energy is lost to friction. 
Therefore, the change in the car's kinetic energy is equal to the work done by the frictional force of 
the car's brakes. 
Therefore, we first need to determine the car's kinetic energy at the moment of braking using: 

KE = -mv'^ (8.32) 

This energy is equal to the work done by the brakes. We have the force applied by the brakes, and 
we can use: 

W = F-d (8.33) 

to determine the stopping distance. 



KE = Imv"^ 



Step 3. 

KE = 

= i(l 000 %) (16,7 ms)'' (8.34) 

139 445 J 
Step 4. Assume the stopping distance is do. Then the work done is: 

W = F-d 

(8.35) 

(-8 000 N) (do) 

The force has a negative sign because it acts in a direction opposite to the direction of motion. 
Step 5. The change in kinetic energy is equal to the work done. 



(8.36) 



AKE 


= 


W 


KEf - KE, 


= 


(-8 000 N) {do) 


J - 139 445 J 


= 


(-8 000 N) (do) 


.•. do 


= 


139 445 J 

8 000 N 




= 


17,4 m 



Step 6. The car stops in 17,4 m. 

Solution to Exercise 8.7 (p. 191) 

Step 1. We are given: 

• we are given the force, F=10 N 

• we are given the speed, v=l m-s~^ 



We are required to calculate the power required. 

Step 2 Itnage notjinished 

Figure 8.21 



Step 4. 



199 

Step 3. From the force diagram, we see that the weight of the box is acting at right angles to the direction 
of motion. The weight does not contribute to the work done and does not contribute to the power 
calculation. 
We can therefore calculate power from: 

P= F -v (8.37) 

P = F -v 

= {10N)(lm-s~^) (8.38) 

WW 

Step 5. 10 W of power are required for a force of 10 N to move a 10 kg box at a speed of 1 ms over a frictionless 
surface. 

Solution to Exercise 8.8 (p. 191) 

Step 1. We are given: 

• mass of crate: m=100 kg 

• height that crate is raised: h=8 m 

• time taken to raise crate: tr=4 s 

• time that crate is held in place: ts=20 s 

We are required to calculate the power exerted. 
Step 2. We can use: 

P = F^ (8.39) 

to calculate power. The force required to raise the crate is equal to the weight of the crate. 
Step 3. 

^ — ^ At 

= rn ■ a — 

^* (8.40) 

= (100kg) (9,8m -5-2^ If 

1 960 W" 

Step 4. While the crate is being held in place, there is no displacement. This means there is no work done on 

the crate and therefore there is no power exerted. 
Step 5. 1 960 W of power is exerted to raise the crate and no power is exerted to hold the crate in place. 



200 CHAPTER 8. WORK, ENERGY AND POWER 



Chapter 9 

Doppler Effect' 



9.1 Introduction 

Have you noticed how the pitch of a poHce car siren changes as the car passes by or how the pitch of a radio 
box on the pavement changes as you drive by? This effect is known as the Doppler Effect and will be 
studied in this chapter. 

NOTE: The Doppler Effect is named after Johann Christian Andreas Doppler (29 November 1803 
- 17 March 1853), an Austrian mathematician and physicist who first explained the phenomenon 
in 1842. 

9.2 The Doppler Effect with Sound and Ultrasound 

As seen in the introduction, there are two situations which lead to the Doppler Effect: 

1. When the source moves relative to the observer, for example the pitch of a car hooter as it passes by. 

2. When the observer moves relative to the source, for example the pitch of a radio on the pavement as 
you drive by. 

Definition 9.1: Doppler Effect 

The Doppler effect is the apparent change in frequency and wavelength of a wave when the observer 
and the source of the wave move relative to each other. 

We experience the Doppler effect quite often in our lives, without realising that it is science taking place. 
The changing sound of a taxi hooter or ambulance as it drives past are examples of this as you have seen in 
the introduction. 

The question is how does the Doppler effect take place. Let us consider a source of sound waves with a 
constant frequency and amplitude. The sound waves can be drawn as concentric circles where each circle 
represents another wavefront, like in Figure 9.1 below. 



Image notjinished 

Figure 9.1: Stationary sound source 



^This content is available online at <http://siyavula.cnx.Org/content/m30847/l.4/>. 

201 



202 CHAPTER 9. DOPPLER EFFECT 

The sound source is the dot in the middle and is stationary. For the Doppler effect to take place, the 
source must be moving relative to the observer. Let's consider the following situation: The source (dot) 
emits one peak (represented by a circle) that moves away from the source at the same rate in all directions. 
The distance between the peaks represents the wavelength of the sound. The closer together the peaks, the 
higher the frequency (or pitch) of the sound. 



Image notjinished 

Figure 9.2 



As this peak moves away, the source also moves and then emits the second peak. Now the two circles are 
not concentric any more, but on the one side they are closer together and on the other side they are further 
apart. This is shown in the next diagram. 



Image notjinished 



Figure 9.3 



If the source continues moving at the same speed in the same direction (i.e. with the same velocity which 
you will learn more about later), then the distance between peaks on the right of the source is constant. The 
distance between peaks on the left is also constant but they are different on the left and right. 



Image notjinished 

Figure 9.4 



This means that the time between peaks on the right is less so the frequency is higher. It is higher than 
on the left and higher than if the source were not moving at all. 

On the left hand side the peaks are further apart than on the right and further apart than if the source 
were at rest - this means the frequency is lower. 

When a car appoaches you, the sound waves that reach you have a shorter wavelength and a higher 
frequency. You hear a higher sound. When the car moves away from you, the sound waves that reach you 
have a longer wavelength and lower frequency. You hear a lower sound. 

This change in frequency can be calculated by using: 

/. = ^/. (9.1) 

where /^ is the frequency perceived by the listener, 
fs is the frequency of the source, 
V is the velocity of the waves, 
vl the velocity of the listener and 



203 

vs the velocity of the source. 

Note: Velocity is a vector and has magnitude and direction. It is very important to get the signs of the 
velocities correct here: 



Source moves towards listener 


Vs : negative 


Source moves away from listener 


Vs : positive 






Listener moves towards source 


vl '■ positive 


Listener moves away from source 


vl '■ negative 



Table 9.1 



Khan academy video on the Doppler effect 



This media object is a Flash object. Please view or download it at 
<http://www.youtube.com/v/dc717Qqa8xk&rel=0&hl=en_US&feature=player_embedded&version=3> 

Figure 9.5 



Exercise 9.1: The Doppler Effect for Sound (Solution on p. 207.) 

The siren of an ambulance has a frequency of 700 Hz. You are standing on the pavement. If the 
ambulance drives past you at a speed of 20 m-s~^, what frequency will you hear, when 

1. the ambulance is approaching you 

2. the ambulance is driving away from you 

Take the speed of sound to be 340 m-s~^. 

Exercise 9.2: The Doppler Effect for Sound 2 (Solution on p. 207.) 

What is the frequency heard by a person driving at 15 m-s^^ toward a factory whistle that is 
blowing at a frequency of 800 Hz. Assume that the speed of sound is 340 m-s~^. 



NOTE: Radar-based speed-traps use the Doppler Effect. The radar gun emits radio waves of a 
specific frequency. When the car is standing still, the waves reflected waves are the same frequency 
as the waves emitted by the radar gun. When the car is moving the Doppler frequency shift can 
be used to determine the speed of the car. 



9.2,1 Ultrasound and the Doppler EfFect 

Ultrasonic waves (ultrasound) are sound waves with a frequency greater than 20 000 Hz (the upper limit of 
human hearing). These waves can be used in medicine to determine the direction of blood flow. The device, 
called a Doppler flow meter, sends out sound waves. The sound waves can travle through skin and tissue and 
will be reflected by moving objects in the body (like blood). The reflected waves return to the flow meter 
where its frequency (received frequency) is compared to the transmitted frequency. Because of the Doppler 
effect, blood that is moving towards the flow meter will change the sound to a higher frequency and blood 
that is moving away from the flow meter will cause a lower frequency. 



204 CHAPTER 9. DOPPLER EFFECT 



Image notjinished 



Figure 9.6 



Ultrasound can be used to determine whether blood is flowing in the right direction in the circulation 
system of unborn babies, or identify areas in the body where blood flow is restricted due to narrow veins. 
The use of ultrasound equipment in medicine is called sonography or ultrasonography. 

9.2.1.1 The Doppler Effect with Sound 

1. Suppose a train is approaching you as you stand on the platform at the station. As the train approaches 
the station, it slows down. All the while, the engineer is sounding the hooter at a constant frequency 
of 400 Hz. Describe the pitch and the changes in pitch that you hear. 

2. Passengers on a train hear its whistle at a frequency of 740 Hz. Anja is standing next to the train 
tracks. What frequency does Anja hear as the train moves directly toward her at a speed of 25 m-s~^? 

3. A small plane is taxiing directly away from you down a runway. The noise of the engine, as the pilot 
hears it, has a frequency 1,15 times the frequency that you hear. What is the speed of the plane? 

9.3 The Doppler Effect with Light 

Light is a wave and earlier you learnt how you can study the properties of one wave and apply the same 
ideas to another wave. The same applies to sound and light. We know the Doppler effect affects sound waves 
when the source is moving. Therefore, if we apply the Doppler effect to light, the frequency of the emitted 
light should change when the source of the light is moving relative to the observer. 

When the frequency of a sound wave changes, the sound you hear changes. When the frequency of light 
changes, the colour you would see changes. 

This means that the Doppler effect can be observed by a change in sound (for sound waves) and a change 
in colour (for light waves). Keep in mind that there are sounds that we cannot hear (for example ultrasound) 
and light that we cannot see (for example ultraviolet light). 

We can apply all the ideas that we learnt about the Doppler effect to light. When talking about light 
we use slightly different names to describe what happens. If you look at the colour spectrum (more details 
Chapter ) then you will see that blue light has shorter wavelengths than red light. If you are in the middle 
of the visible colours then longer wavelengths are more red and shorter wavelengths are more blue. So we 
call shifts towards longer wavelengths "red-shifts" and shifts towards shorter wavelengths "blue-shifts". 



Image notjinished 

Figure 9.7: Blue light has shorter wavelengths than red light. 



A shift in wavelength is the same as a shift in frequency. Longer wavelengths of light have lower frequencies 
and shorter wavelengths have higher frequencies. From the Doppler effect we know that when things move 



205 

towards you any waves they emit that you measure are shifted to shorter wavelengths (blueshifted). If things 
move away from you, the shift is to longer wavelengths (redshifted). 

9.3.1 The Expanding Universe 

Stars emit light, which is why we can see them at night. Galaxies are huge collections of stars. An example is 
our own Galaxy, the Milky Way, of which our sun is only one of the millions of stars! Using large telescopes 
like the Southern African Large Telescope (SALT) in the Karoo, astronomers can measure the light from 
distant galaxies. The spectrum of light can tell us what elements are in the stars in the galaxies because 
each element emits/absorbs light at particular wavelengths (called spectral lines). If these lines are observed 
to be shifted from their usual wavelengths to shorter wavelengths, then the light from the galaxy is said to 
be blueshifted. If the spectral lines are shifted to longer wavelengths, then the light from the galaxy is said 
to be redshifted. If we think of the blueshift and redshift in Doppler effect terms, then a blueshifted galaxy 
would appear to be moving towards us (the observers) and a redshifted galaxy would appear to be moving 
away from us. 

TIP: 

• If the light source is moving away from the observer (positive velocity) then the observed 
frequency is lower and the observed wavelength is greater (redshifted). 

• If the source is moving towards (negative velocity) the observer, the observed frequency is 
higher and the wavelength is shorter (blueshifted). 

Edwin Hubble (20 November 1889 - 28 September 1953) measured the Doppler shift of a large sample 
of galaxies. He found that the light from distant galaxies is redshifted and he discovered that there is a 
proportionality relationship between the redshift and the distance to the galaxy. Galaxies that are further 
away always appear more redshifted than nearby galaxies. Remember that a redshift in Doppler terms means 
a velocity of the light source away from the observer. So why do all distant galaxies appear to be moving 
away from our Galaxy? 

The reason is that the universe is expanding! The galaxies are not actually moving themselves, rather 
the space between them is expanding! 

9.4 Summary 

1. The Doppler Effect is the apparent change in frequency and wavelength of a wave when the observer 
and source of the wave move relative to each other. 

2. The following equation can be used to calculate the frequency of the wave according to the observer 
or listener: 

/. = ^/s (9.2) 

3. If the direction of the wave from the listener to the source is chosen as positive, the velocities have the 
following signs: 



Source moves towards listener 


Vs : negative 


Source moves away from listener 


Vs : positive 






Listener moves towards source 


vl '■ positive 


Listener moves away from source 


vl '■ negative 



206 CHAPTER 9. DOPPLER EFFECT 

Table 9.2 

4. The Doppler Effect can be observed in all types of waves, including ultrasound, light and radiowaves. 

5. Sonography makes use of ultrasound and the Doppler Effect to determine the direction of blood flow. 

6. Light is emitted by stars. Due to the Doppler Effect, the frequency of this light decreases and the 
starts appear red. This is called a red shift and means that the stars are moving away from the Earth. 
This means that the Universe is expanding. 

9.5 End of Chapter Exercises 

1. Write a definition for each of the following terms. 

a. Doppler Effect 

b. Red-shift 

c. Ultrasound 

2. Explain how the Doppler Effect is used to determine the direction of blood flow in veins. 

3. The hooter of an appoaching taxi has a frequency of 500 Hz. If the taxi is travelling at 30 m-s~^and 
the speed of sound is 300 m-s~^, calculate the frequency of sound that you hear when 

a. the taxi is approaching you. 

b. the taxi passed you and is driving away. 

4. A truck approaches you at an unknown speed. The sound of the trucks engine has a frequency of 
210 Hz, however you hear a frequency of 220 Hz. The speed of sound is 340 m-s~^. 

a. Calculate the speed of the truck. 

b. How will the sound change as the truck passes you? Explain this phenomenon in terms of the 
wavelength and frequency of the sound. 

5. A police car is driving towards a fleeing suspect at ^ m.s~^, where v is the speed of sound. The 
frequency of the police car's siren is 400 Hz. The suspect is running away at ^. What frequency does 
the suspect hear? 

6. a. Why are ultrasound waves used in sonography and not sound waves? 

b. Explain how the Doppler effect is used to determine the direction of flow of blood in veins. 



207 



Solutions to Exercises in Chapter 9 

Solution to Exercise 9.1 (p. 203) 

Step 1. 



Step 2. 
Step 3. 

Solution to Exercise 9.2 (p. 203) 
Step 1. We can use 

with: 



V 
Vl 
Vs 
Vs 



h 



V + Vl 

V + Vs 



fs 



700Hz 

340 m-s^i 



-20 TO • s^^ for (a) and 

+20TO-s"ifor (6) 



340+0 C700) 
340-20 \'"") 



743,75Hz 



340+CL (700) 



340+20 

661,11Hz 



V + Vl 

V + Vs' 



V 


= 


340TO-S-1 


VL 


= 


+ 15 TO- s^i 


Vs 


= 


Oto-s^i 


fs 


= 


800 Hz 


II 


= 


? 



The listener is moving towards the source, so vl is positive. 



Step 2. 



v+vl f 

^4°n'^r""'il'n^""'"i' (800 Hz) 

340,6 m-s^+Orri-s^ V '' 

835 Hz 



(9.3) 



(9.4) 



(9.5) 



(9.6) 



(9.7) 



(9.J 



(9.9) 



Step 3. The driver hears a frequency of 835 Hz. 



208 CHAPTER 9. DOPPLER EFFECT 



Chapter 10 
Colour 



10.1 Colour and light' 

10.1.1 Introduction 

The light that human beings can see is called visible light. Visible light is actually just a small part of the large 
spectrum of electromagnetic radiation which you will learn more about in . We can think of electromagnetic 
radiation and visible light as transverse waves. We know that transverse waves can be described by their 
amplitude, frequency (or wavelength) and velocity. The velocity of a wave is given by the product of its 
frequency and wavelength: 

v=fxX (10.1) 

However, electromagnetic radiation, including visible light, is special because, no matter what the frequency, 
it all moves at a constant velocity (in vacuum) which is known as the speed of light. The speed of light 
has the symbol c and is: 

c = 3 X 10^ TO.s-^ (10.2) 

Since the speed of light is c, we can then say: 

c=fx\ (10.3) 



10.1.2 Colour and Light 

Our eyes are sensitive to visible light over a range of wavelengths from 390 nm to 780 nm (1 nm = 1 x 10~^ 
m). The different colours of light we see are related to specific frequencies (and wavelengths) of visible 
light. The wavelengths and frequencies are listed in Table 10.1. 



^This content is available online at <http://siyavula.cnx.Org/content/m39506/l.l/>. 



209 



210 



CHAPTER 10. COLOUR 



Colour 


Wavelength range (nm) 


Frequency range (Hz) 


violet 


390 - 455 


769-659 xl0i2 


blue 


455 - 492 


659-610 xl0i2 


green 


492 - 577 


610- 520 xl0^2 


yellow 


577 - 597 


520- 503 xlO^^ 


orange 


597 - 622 


503- 482 xlO^^ 


red 


622 - 780 


482- 385 xlO^^ 



Table 10.1: Colours, wavelengths and frequencies of light in the visible spectrum. 

You can see from Table 10.1 that violet light has the shortest wavelengths and highest frequencies while 
red light has the longest wavelengths and lowest frequencies. 

Exercise 10.1: Calculating the frequency of light given the wavelength (Solution on p. 

217.) 
A streetlight emits light with a wavelength of 520 nm. 

1. What colour is the light? (Use Table 10.1 to determine the colour) 

2. What is the frequency of the light? 

Exercise 10.2: Calculating the wavelength of light given the frequency (Solution on p. 

217.) 
A streetlight also emits light with a frequency of 490x10^^ Hz. 

1. What colour is the light? (Use Table 10.1 to determine the colour) 

2. What is the wavelength of the light? 

Exercise 10.3: Frequency of Green (Solution on p. 217.) 

The wavelength of green light ranges between 500 nm an d 565 nm. Calculate the range of 
frequencies that correspond to this range of wavelengths. 



10.1.2.1 Calculating w^avelengths and frequencies of light 

1. Calculate the frequency of light which has a wavelength of 400 nm. (Remember to use S.I. units) 

2. Calculate the wavelength of light which has a frequency of 550 x 10^^ Hz. 

3. What colour is light which has a wavelength of 470 x 10^^ m and what is its frequency? 

4. What is the wavelength of light with a frequency of 510 x 10^^ Hz and what is its color? 



10.1.2.2 Dispersion of white light 

White light, like the light which comes from the sun, is made up of all the visible wavelengths of light. In 
other words, white light is a combination of all the colours of visible light. 

You learnt that the speed of light is different in different substances. The speed of light in different 
substances depends on the frequency of the light. For example, when white light travels through glass, 
light of the different frequencies is slowed down by different amounts. The lower the frequency, the less the 
speed is reduced which means that red light (lowest frequency) is slowed down less than violet light (highest 
frequency). We can see this when white light is incident on a glass prism. 



211 

Have a look at the picture below. When the white light hits the edge of the prism, the light which travels 
through the glass is refracted as it moves from the less dense medium (air) to the more dense medium (glass). 



Image notjinished 

Figure 10.1 



• The red light which is slowed down the least, is refracted the least. 

• The violet light which is slowed down the most, is refracted the most. 



When the light hits the other side of the prism it is again refracted but the angle of the prism edge allows 
the light to remain separated into its different colours. White light is therefore separated into its different 
colours by the prism and we say that the white light has been dispersed by the prism. 

The dispersion effect is also responsible for why we see rainbows. When sunlight hits drops of water in 
the atmosphere, the white light is dispersed into its different colours by the water. 

10.1.3 Addition and Subtraction of Light 

10.1.3.1 Additive Primary Colours 

The primary colours of light are red, green and blue. When all the primary colours are superposed (added 
together), white light is produced. Red, green and blue are therefore called the additive primary colours. 
All the other colours can be produced by different combinations of red, green and blue. 

10.1.3.2 Subtractive Primary Colours 

The subtractive primary colours are obtained by subtracting one of the three additive primary colours from 
white light. The subtractive primary colours are yellow, magenta and cyan. Magenta appears as a pinkish- 
purplish colour and cyan looks greenish-blue. You can see how the primary colours of light add up to the 
different subtractive colours in the illustration below. 



Image notjinished 

Figure 10.2 



10.1.3.2.1 Experiment : Colours of light 

Aim: 

To investigate the additive properties of colours and determine the complementary colours of light. 

Apparatus: 

You will need two battery operated torches with fiat bulb fronts, a large piece of white paper, and some 
pieces of cellophane paper of the following colours: red, blue, green, yellow, cyan, magenta. (You should 
easily be able to get these from a newsagents.) 

Make a table in your workbook like the one below: 



212 



CHAPTER 10. COLOUR 



Colour 1 


Colour 2 


Final colour prediction 


Final colour measured 


red 


blue 






red 


green 






green 


blue 






magenta 


green 






yellow 


blue 






cyan 


red 







Table 10.2 

Before you begin your experiment, use what you know about colours of light to write down in the third 
column "Final colour prediction", what you think the result of adding the two colours of light will be. You 
will then be able to test your predictions by making the following measurements: 

Method: 

Proceed according to the table above. Put the correct colour of cellophane paper over each torch bulb, 
e.g. the first test will be to put red cellophane on one torch and blue cellophane on the other. Switch on the 
torch with the red cellophane over it and shine it onto the piece of white paper. 

What colour is the light? 

Turn off that torch and turn on the one with blue cellophane and shine it onto the white paper. 

What colour is the light? 

Now shine both torches with their cellophane coverings onto the same spot on the white paper. What is 
the colour of the light produced? Write this down in the fourth column of your table. 

Repeat the experiment for the other colours of cellophane so that you can complete your table. 

Questions: 

1. How did your predictions match up to your measurements? 

2. Complementary colours of light are defined as the colours of light which, when added to one of the 
primary colours, produce white light. From your completed table, write down the complementary 
colours for red, blue and green. 



10.1.3.3 Complementary Colours 

Complementary colours are two colours of light which add together to give white. 

10.1.3.3.1 Investigation : Complementary colours for red, green and blue 

Complementary colours are two colours which add together to give white. Place a tick in the box where the 
colours in the first column added to the colours in the top row give white. 





magenta 


yellow 


cyan 




(=red+blue) 


(=red+green) 


(=blue+green) 


red 








green 








blue 









Table 10.3 



213 



You should have found that the complementary colours for red, green and blue are: 



Red and Cyan 
Green and Magenta 
Blue and Yellow 



10.1.3.4 Perception of Colour 

The light-sensitive lining on the back inside half of the human eye is called the retina. The retina contains 
two kinds of light sensitive cells or photoreceptors: the rod cells (sensitive to low light) and the cone cells 
(sensitive to normal daylight) which enable us to see. The rods are not sensitive to colour but work well 
in dimly lit conditions. This is why it is possible to see in a dark room, but it is hard to see any colours. 
Only your rods are sensitive to the low light levels and so you can only see in black, white and grey. The 
cones enable us to see colours. Normally, there are three kinds of cones, each containing a different pigment. 
The cones are activated when the pigments absorb light. The three types of cones are sensitive to (i.e. 
absorb) red, blue and green light respectively. Therefore we can perceive all the different colours in the 
visible spectrum when the different types of cones are stimulated by different amounts since they are just 
combinations of the three primary colours of light. 

The rods and cones have different response times to light. The cones react quickly when bright light falls 
on them. The rods take a longer time to react. This is why it takes a while (about 10 minutes) for your eyes 
to adjust when you enter a dark room after being outside on a sunny day. 

NOTE: Color blindness in humans is the inability to perceive differences between some or all 
colors that other people can see. Most often it is a genetic problem, but may also occur because of 
eye, nerve, or brain damage, or due to exposure to certain chemicals. The most common forms of 
human color blindness result from problems with either the middle or long wavelength sensitive cone 
systems, and involve difficulties in discriminating reds, yellows, and greens from one another. This 
is called "red- green color blindness". Other forms of color blindness are much rarer. They include 
problems in discriminating blues from yellows, and the rarest forms of all, complete color blindness 
or monochromasy, where one cannot distinguish any color from grey, as in a black-and-white movie 
or photograph. 



Image notjinished 

Figure 10.3 



run demo^ 

Exercise 10.4: Seeing Colours (Solution on p. 218.) 

When blue and green light fall on an eye, is cyan light being created? Discuss. 



10.1.3.5 Colours on a Television Screen 

K you look very closely at a colour cathode-ray television screen or cathode-ray computer screen, you will 
see that there are very many small red, green and blue dots called phosphors on it. These dots are caused to 



^http://phet. colorado.edu/sims/color- vision/color- visionen.jnlp 



214 CHAPTER 10. COLOUR 

fluoresce (glow brightly) when a beam of electrons from the cathode-ray tube behind the screen hits them. 
Since different combinations of the three primary colours of light can produce any other colour, only red, 
green and blue dots are needed to make pictures containing all the colours of the visible spectrum. 

10.1.3.5.1 Colours of light 

1. List the three primary colours of light. 

2. What is the term for the phenomenon whereby white light is split up into its different colours by a 
prism? 

3. What is meant by the term "complementary colour" of light? 

4. When white light strikes a prism which colour of light is refracted the most and which is refracted the 
least? Explain your answer in terms of the speed of light in a medium. 



10.2 Paints and pigments^ 

10.2.1 Pigments and Paints 

We have learnt that white light is a combination of all the colours of the visible spectrum and that each 
colour of light is related to a different frequency. But what gives everyday objects around us their different 
colours? 

Pigments are substances which give an object its colour by absorbing certain frequencies of light and 
reflecting other frequencies. For example, a red pigment absorbs all colours of light except red which it 
reflects. Paints and inks contain pigments which gives the paints and inks different colours. 

10.2.1.1 Colour of opaque objects 

Objects which you cannot see through (i.e. they are not transparent) are called opaque. Examples of some 
opaque objects are metals, wood and bricks. The colour of an opaque object is determined by the colours 
(therefore frequencies) of light which it reflects. For example, when white light strikes a blue opaque object 
such as a ruler, the ruler will absorb all frequencies of light except blue, which will be reflected. The reflected 
blue light is the light which makes it into our eyes and therefore the object will appear blue. 

Opaque objects which appear white do not absorb any light. They reflect all the frequencies. Black 
opaque objects absorb all frequencies of light. They do not reflect at all and therefore appear to have no 
colour. 

Exercise 10.5: Colour of Opaque Objects (Solution on p. 218.) 

If we shine white light on a sheet of paper that can only reflect green light, what is the colour of 
the paper? 

Exercise 10.6: Colour of an opaque object II (Solution on p. 218.) 

The cover of a book appears to have a magenta colour. What colours of light does it reflect and 
what colours does it absorb? 



10.2.1.2 Colour of transparent objects 

K an object is transparent it means that you can see through it. For example, glass, clean water and some 
clear plastics are transparent. The colour of a transparent object is determined by the colours (frequencies) 
of light which it transmits (allows to pass through it). For example, a cup made of green glass will appear 
green because it absorbs all the other frequencies of light except green, which it transmits. This is the light 
which we receive in our eyes and the object appears green. 



^This content is available online at <http://siyavula.cnx.Org/content/m39505/l.l/>. 



215 



Exercise 10.7: Colour of Transparent Objects (Solution on p. 218.) 

If white light is shone through a glass plate that absorbs light of all frequencies except red, what 
is the colour of the glass plate? 



10.2.1.3 Pigment primary colours 

The primary pigments and paints are cyan, magenta and yellow. When pigments or paints of these three 
colours are mixed together in equal amounts they produce black. Any other colour of paint can be made 
by mixing the primary pigments together in different quantities. The primary pigments are related to the 
primary colours of light in the following way: 



Image notjinished 

Figure 10.4 



NOTE: Colour printers only use 4 colours of ink: cyan, magenta, yellow and black. All the other 
colours can be mixed from these! 

Exercise 10.8: Pigments (Solution on p. 218.) 

What colours of light are absorbed by a green pigment? 

Exercise 10.9: Primary pigments (Solution on p. 218.) 

I have a ruler which reflects red light and absorbs all other colours of light. What colour does the 
ruler appear in white light? What primary pigments must have been mixed to make the pigment 
which gives the ruler its colour? 

Exercise 10.10: Paint Colours (Solution on p. 218.) 

If cyan light shines on a dress that contains a pigment that is capable of absorbing blue, what 
colour does the dress appear? 



This media object is a Flash object. Please view or download it at 

<http://static.slidesharecdn.com/swf/ssplayer2.swf?doc=colour-10051 1051 146- 

phpapp01&stripped_title=colour-4047557&userName=kwarne> 



Figure 10.5 



10.2.2 End of Chapter Exercises 

1. Calculate the wavelength of light which has a frequency of 570 x 10^^ Hz. 

2. Calculate the frequency of light which has a wavelength of 580 nm. 

3. Complete the following sentence: When white light is dispersed by a prism, light of the colour ? 
refracted the most and light of colour ? is refracted the least. 



216 



CHAPTER 10. COLOUR 



4. What are the two types of photoreceptor found in the retina of the human eye called and which type 
is sensitive to colours? 

5. What color do the following shirts appear to the human eye when the lights in a room are turned off 
and the room is completely dark? 

a. red shirt 

b. blue shirt 

c. green shirt 

6. Two light bulbs, each of a different colour, shine on a sheet of white paper. Each light bulb can be a 
primary colour of light - red, green, and blue. Depending on which primary colour of light is used, the 
paper will appear a different color. What colour will the paper appear if the lights are: 

a. red and blue? 

b. red and green? 

c. green and blue? 

7. Match the primary colour of light on the left to its complementary colour on the right: 



Column A 


Column B 


red 


yellow 


green 


cyan 


blue 


magenta 



Table 10.4 

8. Which combination of colours of light gives magenta? 

a. red and yellow 

b. green and red 

c. blue and cyan 

d. blue and red 

9. Which combination of colours of light gives cyan? 

a. yellow and red 

b. green and blue 

c. blue and magenta 

d. blue and red 

10. If yellow light falls on an object whose pigment absorbs green light, what colour will the object appear? 



11. If yellow light falls on a blue pigment, what colour will it appear? 



217 

Solutions to Exercises in Chapter 10 

Solution to Exercise 10.1 (p. 210) 

Step 1. We need to determine the colour and frequency of light with a wavelength of A = 520 nm = 520 x 10"*^ 

ni. 
Step 2. We see from Table 10.1 that light with wavelengths between 492 - 577 nm is green. 520 nm falls into 

this range, therefore the colour of the light is green. 
Step 3. We know that 

c = / X A (10.4) 

We know c and we are given that A = 520 x lO^'' m. So we can substitute in these values and solve 
for the frequency /. (NOTE: Don't forget to always change units into S.I. units! 1 nm = 1 x 10~^ 
m.) 



/ 



A 

3x10'^ 
520x10-^ 

577 X 10^2 H2 



(10.5) 



The frequency of the green light is 577 x 10^^ Hz 
Solution to Exercise 10.2 (p. 210) 

Step 1. We need to find the colour and wavelength of light which has a frequency of 490x10^^ Hz and which 

is emitted by the streetlight. 
Step 2. We can see from Table 10.1 that orange light has frequencies between 503 - 482x10^^ Hz. The light 

from the streetlight has / = 490 x 10^^ Hz which fits into this range. Therefore the light must be 

orange in colour. 
Step 3. We know that 

c = / X A (10.6) 

We know c = 3 x 10® m.s~^ and we are given that / = 490 x 10^^ Hz. So we can substitute in these 
values and solve for the wavelength A. 

^ = 7 

_ 3x10'' 

490x1012 

= 6.122 X 10-7 m (10.7) 

= 612 X 10"^ m 
= 612 nm 

Therefore the orange light has a wavelength of 612 nm. 
Solution to Exercise 10.3 (p. 210) 
Step 1. Use 

c=/xA (10.8) 

to determine /. 



218 CHAPTER 10. COLOUR 

Step 2. 

c = / X A 

■^ " / 1 (10.9) 

_ 3x10'' m-s~^ ^ ' 

565x10-^ m 

= 5,31xl0i''Hz 

Step 3. 

c = / X A 

f = - 

s^ 1 (10.10) 

_ 3x10'^ m-s~^ ^ ' 

500x10-^ m 

= 6, 00x10^'' Hz 

Step 4. The range of frequencies of green light is 5, 31 x 10^"' Hz to 6, 00 x 10^* Hz. 

Solution to Exercise 10.4 (p. 213) 

Step 1. Cyan light is not created when blue and green light fall on the eye. The blue and green receptors are 
stimulated to make the brain believe that cyan light is being created. 

Solution to Exercise 10.5 (p. 214) 

Step 1. Since the colour of an object is determined by that frequency of light that is reflected, the sheet of 
paper will appear green, as this is the only frequency that is reflected. All the other frequencies are 
absorbed by the paper. 

Solution to Exercise 10.6 (p. 214) 

Step 1. We know that magenta is a combination of red and blue primary colours of light. Therefore the object 
must be reflecting blue and red light and absorb green. 

Solution to Exercise 10.7 (p. 214) 

Step 1. Since the colour of an object is determined by that frequency of light that is transmitted, the glass 
plate will appear red, as this is the only frequency that is not absorbed. 

Solution to Exercise 10.8 (p. 215) 

Step 1. If the pigment is green, then green light must be reflected. Therefore, red and blue light are absorbed. 

Solution to Exercise 10.9 (p. 215) 

Step 1. We need to determine the colour of the ruler and the pigments which were mixed to make the colour. 
Step 2. The ruler reflects red light and absorbs all other colours. Therefore the ruler appears to be red. 
Step 3. Red pigment is produced when magenta and yellow pigments are mixed. Therefore magenta and yellow 
pigments were mixed to make the red pigment which gives the ruler its colour. 

Solution to Exercise 10.10 (p. 215) 

Step 1. Cyan light is made up of blue and green light. 

Step 2. If the dress absorbs the blue light then the green light must be reflected, so the dress will appear green! 



Chapter 11 

2D and 3D wavefronts 



11.1 Wavefronts, Huygen's principle and interference' 

11.1.1 Introduction 

You have learnt about the basic principles of reflection and refraction. In this chapter, you will learn about 
phenomena that arise with waves in two and three dimensions: interference and diffraction. 

11.1.2 Wavefronts 

11.1.2.1 Investigation : Wavefronts 

The diagram shows three identical waves being emitted by three point sources. All points marked with the 
same letter are in phase. Join all points with the same letter. 



Image notjtnished 



Figure 11.1 



What type of lines (straight, curved, etc) do you get? How does this compare to the line that joins the 
sources? 

Consider three point sources of waves. If each source emits waves isotropically (i.e. the same in all 
directions) we will get the situation shown in as shown in Figure 11.2. 



Image notjinished 



Figure 11.2: Wavefronts are imaginary lines joining waves that are in phase. In the example, the 
wavefronts (shown by the grey, vertical lines) join all waves at the crest of their cycle. 



^This content is available online at <http://siyavula.cnx.Org/content/m39488/l.l/>. 



219 



220 CHAPTER 11. 2D AND 3D WAVEFRONTS 

We define a wavefront as the imaginary line that joins waves that are in phase. These are indicated by 
the grey, vertical lines in Figure 11.2. The points that are in phase can be peaks, troughs or anything in 
between, it doesn't matter which points you choose as long as they are in phase. 

11.1.3 The Huygens Principle 

Christiaan Huygens described how to determine the path of waves through a medium. 

Definition 11.1: The Huygens Principle 

Each point on a wavefront acts like a point source of circular waves. The waves emitted from these 
point sources interfere to form another wavefront. 

A simple example of the Huygens Principle is to consider the single wavefront in Figure 11.3. 



Image notjinished 



Figure 11.3: A single wavefront at time t acts as a series of point sources of circular waves that interfere 
to give a new wavefront at a time t + At. The process continues and applies to any shape of waveform. 



Exercise 11.1: Application of the Huygens Principle (Solution on p. 232.) 

Given the wavefront. 



Image notjinished 

Figure 11.4 



use the Huygens Principle to determine the wavefront at a later time. 



NOTE: Christiaan Huygens (14 April 1629 - 8 July 1695), was a Dutch mathematician, astronomer 
and physicist; born in The Hague as the son of Constantijn Huygens. He studied law at the 
University of Leiden and the College of Orange in Breda before turning to science. Historians 
commonly associate Huygens with the scientific revolution. Huygens generally receives minor credit 
for his role in the development of modern calculus. He also achieved note for his arguments that 
light consisted of waves; see: wave-particle duality in Chapter . In 1655, he discovered Saturn's 
moon Titan. He also examined Saturn's planetary rings, and in 1656 he discovered that those rings 
consisted of rocks. In the same year he observed and sketched the Orion Nebula. He also discovered 
several interstellar nebulae and some double stars. 



11.1.4 Interference 

Interference occurs when two identical waves pass through the same region of space at the same time 
resulting in a superposition of waves. There are two types of interference which is of interest: constructive 
interference and destructive interference. 



221 

Constructive interference occurs when both waves have a displacement in the same direction, while de- 
structive interference occurs when one wave has a displacement in the opposite direction to the other, thereby 
resulting in a cancellation. When two waves interfere destructively, the resultant absolute displacement of 
the medium is less than in either of the individual displacements. When total destructive interference occurs, 
there is no displacement of the medium. For constructive interference the displacement of the medium is 
greater than the individual displacements. 

Constructive interference is the result of two waves with similar phase overlapping. This means that 
positive parts of one wave tend to overlap with positive parts of the other, and alike for the negative parts. 
When positive is added to positive and negative is added to negative, the net absolute displacement of the 
medium is greater than the displacements of the individual waves. 

Destructive interference, on the other hand, is the result of two waves with non-similar phase (i.e. anti- 
phase) overlapping. This means that the positive parts of one wave tend to align with the negative parts of 
the second wave. When the waves are added together, the positive and negative contributions lead to a net 
absolute displacement of the medium which is less than the absolute displacements of either of the individual 
waves. A place where destructive interference takes place is called a node. 

Waves can interfere at places where there is never a trough and trough or peak and peak or trough and 
peak at the same time. At these places the waves will add together and the resultant displacement will be 
the sum of the two waves but they won't be points of maximum interference. 

Consider the two identical waves shown in the picture below. The wavefronts of the peaks are shown as 
black lines while the wavefronts of the troughs are shown as grey lines. You can see that the black lines cross 
other black lines in many places. This means two peaks are in the same place at the same time so we will 
have constructive interference where the two peaks add together to form a bigger peak. 



Image notjinished 

Figure 11.5 



Two points sources (A and B) radiate identical waves. The wavefronts of the peaks (black lines) and 
troughs (grey lines) are shown. Constructive interference occurs where two black lines intersect or where 
two gray lines intersect. Destructive interference occurs where a black line intersects with a grey line. 

When the grey lines cross other grey lines there are two troughs in the same place at the same time so 
we will have constructive interference where the two troughs add together to form a bigger trough. 

In the case where a grey line crosses a black line we are seeing a trough and peak at the same time. These 
will cancel each other out and the medium will have no displacement at that point. 

• black line + black line = peak + peak = constructive interference 

• grey line + grey line = trough + trough = constructive interference 

• black line + grey line = grey line + black line = peak + trough = trough + peak = destructive 
interference 

On half the picture below, we have marked the constructive interference with a solid black diamond and the 
destructive interference with a hollow diamond. 



Image notjinished 

Figure 11.6 



222 CHAPTER 11. 2D AND 3D WAVEFRONTS 

To see if you understand it, cover up the half we have marked with diamonds and try to work out which 
points are constructive and destructive on the other half of the picture. The two halves are mirror images 
of each other so you can check yourself. 

11.2 Diffraction' 

11.2.1 Diffraction 

One of the most interesting, and also very useful, properties of waves is diffraction. 

Definition 11.2: Diffraction 

Diffraction is the ability of a wave to spread out in wavefronts as the wave passes through a small 
aperture or around a sharp edge. 

11.2.1.1 Diffraction 

Diffraction refers to various phenomena associated with wave propagation, such as the bending, spreading 
and interference of waves emerging from an aperture. It occurs with any type of wave, including sound 
waves, water waves, electromagnetic waves such as light and radio waves. While diffraction always occurs, 
its effects are generally only noticeable for waves where the wavelength is on the order of the feature size of 
the diffracting objects or apertures. 

For example, if two rooms are connected by an open doorway and a sound is produced in a remote corner 
of one of them, a person in the other room will hear the sound as if it originated at the doorway. 



Image notjtnished 



Figure 11.7 



As far as the second room is concerned, the vibrating air in the doorway is the source of the sound. The 
same is true of light passing the edge of an obstacle, but this is not as easily observed because of the short 
wavelength of visible light. 

This means that when waves move through small holes they appear to bend around the sides because 
there are not enough points on the wavefront to form another straight wavefront. This is bending round the 
sides we call diffraction. 

11.2.1.2 Diffraction 

Diffraction effects are more clear for water waves with longer wavelengths. Diffraction can be demonstrated 
by placing small barriers and obstacles in a ripple tank and observing the path of the water waves as 
they encounter the obstacles. The waves are seen to pass around the barrier into the regions behind it; 
subsequently the water behind the barrier is disturbed. The amount of diffraction (the sharpness of the 
bending) increases with increasing wavelength and decreases with decreasing wavelength. In fact, when the 
wavelength of the waves are smaller than the obstacle, no noticeable diffraction occurs. 



^This content is available online at <http://siyavula.cnx.Org/content/m39491/l.l/>. 



223 

11.2.1.3 Experiment : Diffraction 

Water waves in a ripple tank can be used to demonstrate diffraction and interference. 

• Turn on the wave generator so that it produces waves with a high frequency (short wavelength). 

• Place a few obstacles, one at a time, (e.g. a brick or a ruler) in the ripple tank. What happens 
to the wavefronts as they propagate near/past the obstacles? Draw your observations. 

• How does the diffraction change when you change the size of the object? 

• Now turn down the frequency of the wave generator so that it produces waves with longer wavelengths. 

• Place the same obstacles in the ripple tank (one at a time). What happens to the wavefronts as 
they propagate near/past the obstacles? Draw your observations. 

• How does the diffraction change from the higher frequency case? 

• Remove all obstacles from the ripple tank and insert a second wave generator. Turn on both generators 
so that they start at the same time and have the same frequency. 

• What do you notice when the two sets of wavefronts meet each other? 

• Can you identify regions of constructive and destructive interference? 

• Now turn on the generators so that they are out of phase (i.e. start them so that they do not make 
waves at exactly the same time). 

• What do you notice when the two sets of wavefronts meet each other? 

• Can you identify regions of constructive and destructive interference? 

11.2.1.4 Diffraction through a Slit 

When a wave strikes a barrier with a hole only part of the wave can move through the hole. If the hole is 
similar in size to the wavelength of the wave then diffraction occurs. The waves that come through the hole 
no longer looks like a straight wave front. It bends around the edges of the hole. If the hole is small enough 
it acts like a point source of circular waves. 

Now if we allow the wavefront to impinge on a barrier with a hole in it, then only the points on the 
wavefront that move into the hole can continue emitting forward moving waves - but because a lot of the 
wavefront has been removed, the points on the edges of the hole emit waves that bend round the edges. 

Image notjinished 

Figure 11.8 

If you employ Huygens' principle you can see the effect is that the wavefronts are no longer straight lines. 

Image notjinished 

Figure 11.9 



224 CHAPTER 11. 2D AND 3D WAVEFRONTS 

Each point of the slit acts like a point source. If we think about the two point sources on the edges of the 
slit and call them A and B then we can go back to the diagram we had earlier but with some parts blocked 
by the wall. 

Image notjtnished 

Figure 11.10 



If this diagram were showing sound waves then the sound would be louder (constructive interference) in 
some places and quieter (destructive interference) in others. You can start to see that there will be a pattern 
(interference pattern) to the louder and quieter places. If we were studying light waves then the light would 
be brighter in some places than others depending on the interference. 

The intensity (how bright or loud) of the interference pattern for a single narrow slit looks like this: 



Image notjinished 

Figure 11.11 



The picture above shows how the waves add together to form the interference pattern. The peaks 
correspond to places where the waves are adding most intensely and the minima are places where destructive 
interference is taking place. When looking at interference patterns from light the spectrum looks like: 



Image notjinished 



Figure 11.12 



There is a formula we can use to determine where the peaks and minima are in the interference spectrum. 
There will be more than one minimum. There are the same number of minima on either side of the central 
peak and the distances from the first one on each side are the same to the peak. The distances to the peak 
from the second minimum on each side is also the same, in fact the two sides are mirror images of each other. 
We label the first minimum that corresponds to a positive angle from the centre as m = 1 and the first on 
the other side (a negative angle from the centre) as m = — 1, the second set of minima are labelled m = 2 
and m = —2 etc. 



Image notjinished 

Figure 11.13 



225 

The equation for the angle at which the minima occur is given in the definition below: 

Definition 11.3: Interference Minima 

The angle at which the minima in the interference spectrum occur is: 

sine= 11.1 

a 

where 

is the angle to the minimum 

a is the width of the slit 

A is the wavelength of the impinging wavefronts 

m is the order of the mimimum, in = ±1, ±2, ±3, ... 

Exercise 11.2: Diffraction Minimum I (Solution on p. 232.) 

A slit with a width of 2511 nm has red light of wavelength 650 nm impinge on it. The diffracted 
light interferes on a surface. At which angle will the first minimum be? 

Exercise 11.3: Diffraction Minimum II (Solution on p. 232.) 

A slit with a width of 2511 nm has green light of wavelength 532 nm impinge on it. The diffracted 
light interferers on a surface, at what angle will the first minimum be? 

From the formula sin6 = ^^ you can see that a smaller wavelength for the same slit results in a smaller 
angle to the interference minimum. This is something you just saw in the two worked examples. Do a sanity 
check, go back and see if the answer makes sense. Ask yourself which light had the longer wavelength, which 
light had the larger angle and what do you expect for longer wavelengths from the formula. 

Exercise 11.4: Diffraction Minimum III (Solution on p. 233.) 

A slit has a width which is unknown and has green light of wavelength 532 nm impinge on it. 
The diffracted light interferers on a surface, and the first minimum is measure at an angle of 20.77 
degrees? 



Image notjtnished 

Figure 11.14 



run demo 

11.3 Shock waves and sonic booms^ 
11.3.1 Shock Waves and Sonic Booms 

Now we know that the waves move away from the source at the speed of sound. What happens if the source 
moves at the same time as emitting sounds? Once a sound wave has been emitted it is no longer connected 
to the source so if the source moves it doesn't change the way the sound wave is propagating through the 
medium. This means a source can actually catch up to the sound waves it has emitted. 

The speed of sound is very fast in air, about 340 tti • s~^, so if we want to talk about a source catching 
up to sound waves then the source has to be able to move very fast. A good source of sound waves to discuss 
is a jet aircraft. Fighter jets can move very fast and they are very noisy so they are a good source of sound 
for our discussion. Here are the speeds for a selection of aircraft that can fiy faster than the speed of sound. 

^http://phet. Colorado. edu/sims/wave-interference/wave-interferenceen.jnlp 

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226 



CHAPTER 11. 2D AND 3D WAVEFRONTS 



Aircraft 


speed at altitude (km ■ h ^) 


speed at altitude {m ■ s ^) 


Concorde 


2 330 


647 


Gripen 


2 410 


669 


Mirage Fl 


2 573 


715 


Mig 27 


1885 


524 


F 15 


2 660 


739 


F 16 


2 414 


671 



Table 11.1 



11.3.1.1 Subsonic Flight 

Definition 11.4: Subsonic 

Subsonic refers to speeds slower than the speed of sound. 

When a source emits sound waves and is moving slower than the speed of sound you get the situation in 
this picture. Notice that the source moving means that the wavefronts, and therefore peaks in the wave, are 
actually closer together in the one direction and further apart in the other. 



Image notjinished 



Figure 11.15 



If you measure the waves on the side where the peaks are closer together you'll measure a different 
wavelength than on the other side of the source. This means that the noise from the source will sound 
different on the different sides. This is called the Doppler Effect. 

Definition 11.5: Doppler Effect 

when the wavelength and frequency measured by an observer are different to those emitted by the 
source due to movement of the source or observer. 

11.3.1.2 Supersonic Flight 

Definition 11.6: Supersonic 

Supersonic refers to speeds faster than the speed of sound. 

If a plane flies at exactly the speed of sound then the waves that it emits in the direction it is flying won't 
be able to get away from the plane. It also means that the next sound wave emitted will be exactly on top 
of the previous one, look at this picture to see what the wavefronts would look like: 



Image notjinished 



Figure 11.16 



227 

Sometimes we use the speed of sound as a reference to describe the speed of the object (aircraft in our 
discussion). 

Definition 11.7: Mach Number 

The Mach Number is the ratio of the speed of an object to the speed of sound in the surrounding 
medium. 

Mach number is tells you how many times faster than sound the aircraft is moving. 

• Mach Number < 1 : aircraft moving slower than the speed of sound 

• Mach Number = 1 : aircraft moving at the speed of sound 

• Mach Number > 1 : aircraft moving faster than the speed of sound 

To work out the Mach Number divide the speed of the aircraft by the speed of sound. 

Mach Number = !^2:ll2^ (11.2) 

^sound 

Remember: the units must be the same before you divide. 

If the aircraft is moving faster than the speed of sound then the wavefronts look like this: 



Image notjinished 

Figure 11.17 



If the source moves faster than the speed of sound, a cone of wave fronts is created. This is called a Mach 
cone. From constructive interference, we know that two peaks that add together form a larger peak. In a 
Mach cone many, many peaks add together to form a very large peak. This is a sound wave so the large 
peak is a very, very loud sound wave. This sounds like a huge "boom" and we call the noise a sonic boom. 

Definition 11.8: Sonic Boom 

A sonic boom is the sound heard by an observer as a Shockwave passes. 

Exercise 11.5: Mach Speed I (Solution on p. 233.) 

An aircraft flies at 1300 km • h~ and the speed of sound in air is 340 m • s~^. What is the Mach 
Number of the aircraft? 



11.3.1.2.1 Mach Number 

In this exercise we will determine the Mach Number for the different aircraft in the previous table. To help you 
get started we have calculated the Mach Number for the Concord with a speed of sound v sound = 340 ms~^. 
For the Condorde we know the speed and we know that: 

Mach Number = ^^2:112^ (11.3) 

^sound 

For the Concorde this means that 

Mach Number = ||^ 

34° 11.4 

= 1.9 



228 



CHAPTER 11. 2D AND 3D WAVEFRONTS 



Aircraft 


speed at altitude (km ■ h ^) 


speed at altitude {m ■ s ^) 


Mach Number 


Concorde 


2 330 


647 


1.9 


Gripen 


2 410 


669 




Mirage Fl 


2 573 


715 




Mig27 


1 885 


524 




F 15 


2 660 


739 




F 16 


2 414 


671 





Table 11.2 

Now calculate the Mach Numbers for the other aircraft in the table. 

11.3.1.3 Mach Cone 

The shape of the Mach Cone depends on the speed of the aircraft. When the Mach Number is 1 there is no 
cone but as the aircraft goes faster and faster the angle of the cone gets smaller and smaller. 
If we go back to the supersonic picture we can work out what the angle of the cone must be. 



Image notjtnished 



Figure 11.18 



We build a triangle between how far the plane has moved and how far a wavefront at right angles to the 
direction the plane is flying has moved: 

An aircraft emits a sound wavefront. The wavefront moves at the speed of sound 340 m ■ s~^ and the 
aircraft moves at Mach 1.5, which is 1.5 x 340 = 510 m • s~^. The aircraft travels faster than the wavefront. 
K we let the wavefront travel for a time t then the following diagram will apply: 



Image notjinished 



Figure 11.19 



We know how fast the wavefront and the aircraft are moving so we know the distances that they have 
traveled: 



Image notjinished 



Figure 11.20 



229 

The angle between the cone that forms and the direction of the plane can be found from the right-angle 
triangle we have drawn into the figure. We know that sind = h°^ot°cnusG '^hich in this figure means: 



sinO 
sinO 
sinO 



opposite 
hypotenuse 

"aircra/tX* 
Vsov.nd 



(11.5) 



In this case we have used sound and aircraft but a more general way of saying this is: 



• aircraft = source 

• sound = wavefront 



We often just write the equation as: 



'^aircraft 



sinO 
sin6 



Vg 



VgSinO 



Vairavaft 

'^ sound 

'^wavefront 

Vw 



(11.6) 



From this equation, we can see that the faster the source (aircraft) moves, the smaller the angle of the Mach 
cone. 

11.3.1.3.1 Mach Cone 

In this exercise we will determine the Mach Cone Angle for the different aircraft in the table mentioned 
above. To help you get started we have calculated the Mach Cone Angle for the Concorde with a speed of 
sound V sound = 340 m ■ s~^ . 

For the Condorde we know the speed and we know that: 



For the Concorde this means that 



sinO ■■ 



V, 



aircraft 



(11.7) 



sinO 



340 

647 
T-1340 
^ 647 



(ll.J 



31.7 



O 



Aircraft 


speed at altitude (km ■ h ^) 


speed at altitude {m ■ s ^) 


Mach Cone Angle (degrees) 


Concorde 


2 330 


647 


31.7 


Grip en 


2 410 


669 




Mirage Fl 


2 573 


715 




Mig 27 


1 885 


524 




F 15 


2 660 


739 




F 16 


2 414 


671 





Table 11.3 

Now calculate the Mach Cone Angles for the other aircraft in the table. 



230 CHAPTER 11. 2D AND 3D WAVEFRONTS 

11,3.2 End of Chapter Exercises 

1. In the diagram below the peaks of wavefronts are shown by black lines and the troughs by grey lines. 
Mark all the points where constructive interference between two waves is taking place and where 
destructive interference is taking place. Also note whether the interference results in a peak or a 
trough. 

Image notjinished 

Figure 11.21 



2. For a slit of width 1300 nm, calculate the first 3 minima for light of the following wavelengths: 

a. blue at 475 nm 

b. green at 510 nm 

c. yellow at 570 nm 

d. red at 650 nm 

3. For light of wavelength 540 nm, determine what the width of the slit needs to be to have the first 
minimum at: 

a. 7.76 degrees 

b. 12.47 degrees 

c. 21.1 degrees 

4. For light of wavelength 635 nm, determine what the width of the slit needs to be to have the second 
minimum at: 

a. 12.22 degrees 

b. 18.51 degrees 

c. 30.53 degrees 

5. If the first minimum is at 8.21 degrees and the second minimum is at 16.6 degrees, what is the 
wavelength of light and the width of the slit? (Hint: solve simultaneously.) 

6. Determine the Mach Number, with a speed of sound of 340 m ■ s~^, for the following aircraft speeds: 

a. 640 m ■ s~^ 

b. 980 m ■ s-i 

c. 500 m ■ s~^ 

d. 450 m ■ s~^ 

e. 1300 km-h^^ 

f. 1450 km-h^^ 

g. 1760 km-h^^ 



7. If an aircraft has a Mach Number of 3.3 and the speed of sound is 340 m ■ s ^, what is its speed? 

8. Determine the Mach Cone angle, with a speed of sound of 340 m ■ s~^, for the following aircraft speeds: 



a. 


640 m ■ s- 


-1 




b. 


980 m ■ s' 


~i 




c. 


500 m ■ s' 


-1 




d. 


450 m ■ s' 


-1 




e. 


1300 km • 


h" 


-1 


f. 


1450 km • 


h' 


-1 


g- 


1760 km • 


h^ 


-1 



231 

9. Determine the aircraft speed, with a speed of sound of 340 m ■ s~^, for the following Mach Cone Angles: 

a. 58.21 degrees 

b. 49.07 degrees 

c. 45.1 degrees 

d. 39.46 degrees 

e. 31.54 degrees 



232 CHAPTER 11. 2D AND 3D WAVEFRONTS 

Solutions to Exercises in Chapter 11 

Solution to Exercise 11.1 (p. 220) 



Image notjtnished 



Step 1. 

Figure 11.22 



Image notjinished 



Step 2. 

Figure 11.23 



Solution to Exercise 11.2 (p. 225) 

Step 1. We know that we are dealing with interference patterns from the diffraction of light passing through a 
slit. The slit has a width of 2511 nm which is 2511 x 10^^ m and we know that the wavelength of the 
light is 650 nm which is 650 x 10^^ m. We are looking to determine the angle to first minimum so we 
know that m = 1. 

Step 2. We know that there is a relationship between the slit width, wavelength and interference minimum 
angles: 

sine = 11.9 

a 

We can use this relationship to find the angle to the minimum by substituting what we know and 

solving for the angle. 

Step 3. 

sin9 
sin9 
sine = 0.258861012 (11.10) 

e 

e = 15° 

The first minimum is at 15 degrees from the centre peak. 
Solution to Exercise 11.3 (p. 225) 

Step 1. We know that we are dealing with interference patterns from the diffraction of light passing through a 
slit. The slit has a width of 2511 nm which is 2511 x 10~^ m and we know that the wavelength of the 
light is 532 nm which is 532 x 10~^ m. We are looking to determine the angle to first minimum so we 
know that m = 1. 

Step 2. We know that there is a relationship between the slit width, wavelength and interference minimum 
angles: 

"mX 

sine= (11-11) 

a 

We can use this relationship to find the angle to the minimum by substituting what we know and 

solving for the angle. 



650xl0-^m 


2511xl0-!'m 


650 
2511 


0.258861012 


sm^iO.258861012 



532x10" 


"m 


2511x10- 
532 
2511 


■•■^m 



233 

Step 3. 

sinO = 

sinO = 

sine = 0.211867782 (11-12) 

e = sm^iO.211867782 

e = 12.2° 

The first minimum is at 12.2 degrees from the centre peak. 
Solution to Exercise 11.4 (p. 225) 

Step 1. We know that we are dealing with interference patterns from the diffraction of light passing through a 
slit. We know that the wavelength of the light is 532 nm which is 532 x 10~^ m. We know the angle 
to first minimum so we know that m = 1 and = 20.77°. 

Step 2. We know that there is a relationship between the slit width, wavelength and interference minimum 
angles: 

sine = 11.13 

a 

We can use this relationship to find the width by substituting what we know and solving for the width. 

Step 3. 



sinO 


= 


532x10 " m 
a 


sin20.77° 


= 


532x10"" 
a 


a 


= 


532x10"" 
0.354666667 


a 


= 


1500 X 10"^ 


a 


= 


1500 nm 



(11.14) 



The slit width is 1500 nm. 

Solution to Exercise 11.5 (p. 227) 

Step 1. We know we are dealing with Mach Number. We are given the speed of sound in air, 340 m ■ s"^, and 
the speed of the aircraft, 1300 km • /i~^. The speed of the aircraft is in different units to the speed of 
sound so we need to convert the units: 



1300km • /i"^ = 1300km • /i"^ 
-1 = 1300 X i» 

3d00s 

1300km -/i"^ = 361.1 m-fi-i 



1300km • /i~i = 1300 X »^ (11.15) 



Step 2. We know that there is a relationship between the Mach Number, the speed of sound and the speed of 
the aircraft: 

Mach Number = !^2:1^£^ (11.16) 

^sound 

We can use this relationship to find the Mach Number. 
Step 3. 

Mach Number = f-riraft 



Mach Number = ^ (11-17) 



Mach Number = 1.06 
The Mach Number is 1.06. 



234 CHAPTER 11. 2D AND 3D WAVEFRONTS 



Chapter 12 

Wave nature of matter 



12.1 de Broglie wavelength' 

12.1.1 Introduction 

In chapters and the so-called wave-particle duality of light is described. This duality states that light displays 
properties of both waves and of particles, depending on the experiment performed. For example, interference 
and diffraction of light are properties of its wave nature, while the photoelectric effect is a property of its 
particle nature. In fact we call a particle of light a photon. 

Hopefully you have realised that nature loves symmetry. So, if light which was originally believed to be 
a wave also has a particle nature, then perhaps particles, also display a wave nature. In other words matter 
which we originally thought of as particles may also display a wave-particle duality. 

12.1.2 de Broglie Wavelength 

Einstein showed that for a photon, its momentum, p, is equal to its energy, E divided by the speed of light. 



c: 

p=-. (12.1) 

c 

The energy of the photon can also be expressed in terms of the wavelength of the light. A: 

E=^, (12.2) 

where h is Planck's constant. Combining these two equations we find that the the momentum of the photon 
is related to its wavelength 

he h , ^ 

or equivalently 

A=-. (12.4) 

P 

In 1923, Louis de Broglie proposed that this equation not only holds for photons, but also holds for particles 
of matter. This is known as the de Broglie hypothesis. 



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236 CHAPTER 12. WAVE NATURE OF MATTER 

Definition 12.1: De Broglie Hypothesis 

A particle of mass m moving with velocity v has a wavelength A related to is momentum p = mv 
by 

h h , ^ 

\=- = (12.5) 

p mv 

This wavelength, A, is known as the de Broglie wavelength of the particle (where h is Planck's 
constant). 

Since the value of Planck's constant is incredibly small h = 6.63 x 10^"^^ J • s, the wavelike nature of 
everyday objects is not really observable. 

NOTE: The de Broglie hypothesis was proposed by French physicist Louis de Broglie (15 August 
1892 - 19 March 1987) in 1923 in his PhD thesis. He was awarded the Nobel Prize for Physics in 
1929 for this work, which made him the first person to receive a Nobel Prize on a PhD thesis. 

Exercise 12.1: de Broglie Wavelength of a Cricket Ball (Solution on p. 240.) 

A cricket ball has a mass of 0, 150kg and is bowled towards a bowler at 40 m • s~^. Calculate the 
de Broglie wavelength of the cricket ball? 

This wavelength is considerably smaller than the diameter of a proton which is approximately 10~^^ m. 
Hence the wave-like properties of this cricket ball are too small to be observed. 

Exercise 12.2: The de Broglie wavelength of an electron (Solution on p. 240.) 

Calculate the de Broglie wavelength of an electron moving at 40 m-s~^. 

Although the electron and cricket ball in the two previous examples are travelling at the same velocity the de 
Broglie wavelength of the electron is much larger than that of the cricket ball. This is because the wavelength 
is inversely proportional to the mass of the particle. 

Exercise 12.3: The de Broglie wavelength of an electron (Solution on p. 240.) 

Calculate the de Broglie wavelength of a electron moving at 3 x 10^ m • s~^. {j^ of the speed of 

light.) 
Since the de Broglie wavelength of a particle is inversely proportional to its velocity, the wavelength decreases 
as the velocity increases. This is confirmed in the last two examples with the electrons. De Broglie's 
hypothesis was confirmed by Davisson and Germer in 1927 when they observed a beam of electrons being 
diffracted off a nickel surface. The diffraction means that the moving electrons have a wave nature. They 
were also able to determine the wavelength of the electrons from the diffraction. To measure a wavelength 
one needs two or more diffracting centres such as pinholes, slits or atoms. For diffraction to occur the centres 
must be separated by a distance about the same size as the wavelength. Theoretically, all objects, not just 
sub-atomic particles, exhibit wave properties according to the de Broglie hypothesis. 

Image notjinished 

Figure 12.1: The wavelengths of the fast electrons are much smaller than that of visible light. 



237 

12.2 Electron microscopes' 
12.2.1 The Electron Microscope 

We have seen that under certain circumstances particles behave Hke waves. This idea is used in the electron 
microscope which is a type of microscope that uses electrons to create an image of the target. It has much 
higher magnification or resolving power than a normal light microscope, up to two million times, allowing it 
to see smaller objects and details. 

Let's first review how a regular optical microscope works. A beam of light is shone through a thin target 
and the image is then magnified and focused using objective and ocular lenses. The amount of light which 
passes through the target depends on the densities of the target since the less dense regions allow more light 
to pass through than the denser regions. This means that the beam of light which is partially transmitted 
through the target carries information about the inner structure of the target. 

The original form of the electron microscopy, transmission electron microscopy, works in a similar manner 
using electrons. In the electron microscope, electrons which are emitted by a cathode are formed into a beam 
using magnetic lenses. This electron beam is then passed through a very thin target. Again, the regions in the 
target with higher densities stop the electrons more easily. So, the number of electrons which pass through 
the different regions of the target depends on their densities. This means that the partially transmitted 
beam of electrons carries information about the densities of the inner structure of the target. The spatial 
variation in this information (the "image") is then magnified by a series of magnetic lenses and it is recorded 
by hitting a fluorescent screen, photographic plate, or light sensitive sensor such as a CCD (charge-coupled 
device) camera. The image detected by the CCD may be displayed in real time on a monitor or computer. 
In Figure 12.2 is an image of the polio virus obtained with a transmission electron microscope. 

Image notjinished 

Figure 12.2: The image of the polio virus using a transmission electron microscope. 



The structure of an optical and electron microscope are compared in Figure 12.3. While the optical 
microscope uses light and focuses using lenses, the electron microscope uses electrons and focuses using 
electromagnets. 

Image notjinished 

Figure 12.3: Diagram of the basic components of an optical microscope and an electron microscope. 



Electron microscopes are very useful as they are able to magnify objects to a much higher resolution. 
This is because their de Broglie wavelengths are so much smaller than that of visible light. You hopefully 
remember that light is diffracted by objects which are separated by a distance of about the same size as the 
wavelength of the light. This diffraction then prevents you from being able to focus the transmitted light 



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238 



CHAPTER 12. WAVE NATURE OF MATTER 



into an image. So the sizes at which diffraction occurs for a beam of electrons is much smaller than those for 
visible light. This is why you can magnify targets to a much higher order of magnification using electrons 
rather than visible light. 





Light microscope 


Electron microscope 


Source 


Bright lamp or laser 


Electron gun 


Radiation 


U.V. or visible Hght 


Electron beam produced by heat- 
ing metal surface (e.g. tungsten) 


Lenses 


Curved glass surfaces 


Electromagnets 


Receiver 


Eye; photographic emulsion or 
digital image 


Fluorescent screen (for location 
and focusing image); photo- 
graphic emulsion or digital image 


Focus 


Axial movement of lenses (up and 
down) 


Adjustment of magnetic field in 
the electromagnets by changing 
the current 


Operating 


Atmospheric 


High vacuum 


Pressure 







Table 12.1: Comparison of Light and Electron Microscopes 



12.2.1.1 High-Resolution Transmission Electron Microscope (HRTEM) 

There are high-resolution TEM (HRTEM) which have been built. In fact the resolution is sufficient to show 
carbon atoms in diamond separated by only 89 picometers and atoms in silicon at 78 picometers. This is at 
magnifications of 50 million times. The ability to determine the positions of atoms within materials has made 
the HRTEM a very useful tool for nano-technologies research. It is also very important for the development 
of semiconductor devices for electronics and photonics. 

Transmission electron microscopes produce two-dimensional images. 

12.2.1.2 Scanning Electron Microscope (SEM) 

The Scanning Electron Microscope (SEM) produces images by hitting the target with a primary electron 
beam which then excites the surface of the target. This causes secondary electrons to be emitted from the 
surface which are then detected. So the electron beam in the SEM is moved (or scanned) across the sample, 
while detectors build an image from the secondary electrons. 

Generally, the transmission electron microscope's resolution is about an order of magnitude better than 
the SEM resolution. However, because the SEM image relies on surface processes rather than transmission 
it is able to image bulk samples (unlike optical microscopes and TEM which require the samples to be thin) 
and has a much greater depth of view, and so can produce images that are a good representation of the 3D 
structure of the sample. 



12.2.1.3 Disadvantages of an Electron Microscope 

Electron microscopes are expensive to buy and maintain. They are also very sensitive to vibration and 
external magnetic fields. This means that special facilities are required to house microscopes aimed at 
achieving high resolutions. Also the targets have to be viewed in vacuum, as the electrons would scatter off 
the molecules that make up air. 



239 

12.2.1.3.1 Scanning Electron Microscope (SEM) 

Scanning electron microscopes usually image conductive or semi-conductive materials best. A common 
preparation technique is to coat the target with a several-nanometer layer of conductive material, such as 
gold, from a sputtering machine; however this process has the potential to disturb delicate samples. 

The targets have to be prepared in many ways to give proper detail. This may result in artifacts purely 
as a result of the treatment. This gives the problem of distinguishing artifacts from material, particularly 
in biological samples. Scientists maintain that the results from various preparation techniques have been 
compared, and as there is no reason that they should all produce similar artifacts, it is therefore reasonable 
to believe that electron microscopy features correlate with living cells. 

NOTE: The first electron microscope prototype was built in 1931 by the German engineers Ernst 
Ruska and Max Knoll. It was based on the ideas and discoveries of Louis de Broglie. Although it 
was primitive and was not ideal for practical use, the instrument was still capable of magnifying 
objects by four hundred times. The first practical electron microscope was built at the University of 
Toronto in 1938, by Eli Franklin Burton and students Cecil Hall, James Hillier and Albert Prebus. 

Although modern electron microscopes can magnify objects up to two million times, they are still 
based upon Ruska's prototype and his correlation between wavelength and resolution. The electron 
microscope is an integral part of many laboratories. Researchers use it to examine biological 
materials (such as microorganisms and cells), a variety of large molecules, medical biopsy samples, 
metals and crystalline structures, and the characteristics of various surfaces. 



12.2.1.4 Uses of Electron Microscopes 

Electron microscopes can be used to study: 

• the topography of an object — how its surface looks. 

• the morphology of particles making up an object — their shapes and sizes. 

• the composition of an object — the elements and compounds that the object is composed of and the 
relative amounts of them. 

• the crystallographic information for crystalline samples — how the atoms are arranged in the object. 



12.2.2 End of Chapter Exercises 

1. If the following particles have the same velocity, which has the shortest wavelength: electron, hydrogen 
atom, lead atom? 

2. A bullet weighing 30 g is fired at a velocity of 500 m • s^^. What is its wavelength? 

3. Calculate the wavelength of an electron which has a kinetic energy of 1.602 x 10~^^ J. 

4. If the wavelength of an electron is 10~^ m what is its velocity? 

5. Considering how one calculates wavelength using slits, try to explain why we would not be able to 
physically observe diffraction of the cricket ball in the first worked example. 



240 CHAPTER 12. WAVE NATURE OF MATTER 

Solutions to Exercises in Chapter 12 

Solution to Exercise 12.1 (p. 236) 

Step 1. We are required to calculate the de Broglie wavelength of a cricket ball given its mass and speed. We 
can do this by using: 



h 

A= (12.6) 

mv 



Step 2. We are given: 



• The mass of the cricket ball m = 0, 150 kg 

• The velocity of the cricket ball v = AOm ■ s~^ 

and we know: 

• Planck's constant h = 6,63 x 10""^"* J • s 
Step 3. 





h 
mv 
6,63x10"^' 


' Js 


(0, 

1 


150 kg) (40 

,11 X 10' 





A 

mv 

(12.7) 



Solution to Exercise 12.2 (p. 236) 

Step 1. We are required to calculate the de Broglie wavelength of an electron given its speed. We can do this 
by using: 

A= (12.8) 

mv 

Step 2. We are given: 

• The velocity of the electron v = 40 m ■ s~^ 
and we know: 

• The mass of the electron m.^ = 9, 11 x 10^^^ kg 

• Planck's constant h = 6,63 x lO^'^* J • s 

Step 3. 

A = ^ 

mv 
6,63x10"-''*' J-s 



(9,llxl0--''i kg)(40 m-s-i) ('12 91 

1,82 X 10"^ m 
0,0182 mm 



Solution to Exercise 12.3 (p. 236) 



Step 1. We are required to calculate the de Broglie wavelength of an electron given its speed. We can do this 
by using: 

h 
A= (12.10) 

m,v 

Step 2. We are given: 

• The velocity of the electron v = 3 x 10^ m • s~^ 



241 

and we know: 

• The mass of the electron m = 9, 11 x 10^"^^ kg 

• Planck's constant /i = 6, 63 x 10"'^'' J • s 



Step 3. 



h 

mv 

6,63x10"^" J-s 

(9,11x10-31 kg)(3xl0= m-s-i) 

2,43 X IQ-'^m 



(12.11) 



This is the size of an atom. For this reason, electrons moving at high velocities can be used to "probe" 
the structure of atoms. This is discussed in more detail at the end of this chapter. Figure 12.1 compares 
the wavelengths of fast moving electrons to the wavelengths of visible light. 



242 CHAPTER 12. WAVE NATURE OF MATTER 



Chapter 13 

Electrodynamics 



13.1 Generators and motors' 

13.1.1 Introduction 

In Grade 11 you learnt how a magnetic field is generated around a current carrying conductor. You also 
learnt how a current is generated in a conductor that moves in a magnetic field. This chapter describes how 
conductors moving in a magnetic field are applied in the real-world. 

13.1.2 Electrical machines - generators and motors 

We have seen that when a conductor is moved in a magnetic field or when a magnet is moved near a 
conductor, a current fiows in the conductor. The amount of current depends on the speed at which the 
conductor experiences a changing magnetic field, the number of turns of the conductor and the position of 
the plane of the conductor with respect to the magnetic field. The effect of the orientation of the conductor 
with respect to the magnetic field is illustrated in Figure 13.1. 



Image notjinished 



Figure 13.1: Series of figures showing that the magnetic flux through a conductor is dependent on the 
angle that the plane of the conductor makes with the magnetic fleld. The greatest flux passes through 
the conductor when the plane of the conductor is perpendicular to the magnetic fleld lines as in (a). The 
number of fleld lines passing through the conductor decreases, as the conductor rotates until it is parallel 
to the magnetic fleld (c). 



If the current fiowing in the conductor were plotted as a function of the angle between the plane of 
the conductor and the magnetic field, then the current would vary as shown in Figure 13.2. The current 
alternates about zero and is known as an alternating current (abbreviated AC). 



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243 



244 CHAPTER 13. ELECTRODYNAMICS 



Image notjinished 



Figure 13.2: Variation of current as the angle of the plane of a conductor with the magnetic field 
changes. 



Recall Faraday's Law which you learnt about in Grade 11: 

Definition 13.1: Faraday's Law 

The emf (electromotive force), e, produced around a loop of conductor is proportional to the rate of 
change of the magnetic flux, (p, through the area, A, of the loop. This can be stated mathematically 
as: 

e = -N^ (13.1) 

At ^ ' 

where <j) = B ■ A and B is the strength of the magnetic field. 

Faraday's Law relates induced emf (electromotive force) to the rate of change of flux, which is the product 
of the magnetic field and the cross-sectional area the field lines pass through. As the closed loop conductor 
changes orientation with respect to the magnetic field, the amount of magnetic flux through the area of the 
loop changes, and an emf is induced in the conducting loop. 

13.1.2.1 Electrical generators 

13.1.2.1.1 AC generator 

The principle of rotating a conductor in a magnetic field is used in electrictrical generators. A generator 
converts mechanical energy (motion) into electrical energy. 

Definition 13.2: Generator 

A generator converts mechanical energy into electrical energy. 

The layout of a simple AC generator is shown in Figure 13.3. The conductor in the shape of a coil 
is connected to a slip ring. The conductor is then manually rotated in the magnetic field generating an 
alternating emf. The slip rings are connected to the load via brushes. 



Image notjinished 

Figure 13.3: Layout of an alternating current generator. 



If a machine is constructed to rotate a magnetic field around a set of stationary wire coils with the turning 
of a shaft, AC voltage will be produced across the wire coils as that shaft is rotated, in accordance with 
Faraday's Law of electromagnetic induction. This is the basic operating principle of an AC generator. 

In an AC generator the two ends of the coil are each attached to a slip ring that makes contact with 
brushes as the coil turns. The direction of the current changes with every half turn of the coil. As one side 
of the loop moves to the other pole of the magnetic field, the current in the loop changes direction. The two 



245 

slip rings of the AC generator allow the coil to turn without breaking the connections to the load circuit. 
This type of current which changes direction is known as alternating current. 

NOTE: AC generators are also known as alternators. They are found in motor cars to charge the 
car battery. 



13.1.2.1.2 DC generator 

A simple DC generator is constructed the same way as an AC generator except that there is one slip ring 
which is split into two pieces, called a commutator, so the current in the external circuit does not change 
direction. The layout of a DC generator is shown in Figure 13.4. The split-ring commutator accommodates 
for the change in direction of the current in the loop, thus creating direct current (DC) current going through 
the brushes and out to the circuit. 



Image notjinished 

Figure 13.4: Layout of a direct current generator. 



The shape of the emf from a DC generator is shown in Figure 13.5. The emf is not steady but is the 
absolute value of a sine/cosine wave. 

Image notjinished 

Figure 13.5: Variation of emf in a DC generator. 

13.1.2.1.3 AC versus DC generators 

The problems involved with making and breaking electrical contact with a moving coil are sparking and heat, 
especially if the generator is turning at high speed. If the atmosphere surrounding the machine contains 
flammable or explosive vapors, the practical problems of spark-producing brush contacts are even greater. 

If the magnetic field, rather than the coil/conductor is rotated, then brushes are not needed in an AC 
generator (alternator), so an alternator will not have the same problems as DC generators. The same benefits 
of AC over DC for generator design also apply to electric motors. While DC motors need brushes to make 
electrical contact with moving coils of wire, AC motors do not. In fact, AC and DC motor designs are very 
similar to their generator counterparts. The AC motor is depends on the reversing magnetic field produced 
by alternating current through its stationary coils of wire to make the magnet rotate. The DC motor depends 
on the brush contacts making and breaking connections to reverse current through the rotating coil every 
1/2 rotation (180 degrees). 



246 CHAPTER 13. ELECTRODYNAMICS 

13.1.2.2 Electric motors 

The basic principles of operation for a motor are the same as that of a generator, except that a motor 
converts electrical energy into mechanical energy (motion). 

Definition 13.3: Motor 

An electric motor converts electrical energy into mechanical energy. 

If one were to place a moving charged particle in a magnetic field, it would feel a force called the Lorentz 
force. 

Definition 13.4: The Lorentz Force 

The Lorentz force is the force experienced by a moving charged particle in a magnetic field and 
can be described by: 

F = qxvxB (13.2) 

where 

F is the force (in newtons, N) 

q is the electric charge (in coulombs, C) 

V is the velocity of the charged particle (in m.s~^) and 

B is the magnetic field strength (in teslas, T). 

Current in a conductor consists of moving charges. Therefore, a current carrying coil in a magnetic field 
will also feel the Lorentz force. For a straight current carrying wire which is not moving: 

F = I X Lx B (13.3) 

where 

F is the force (in newtons, N) 

/ is the current in the wire (in amperes. A) 

L is the length of the wire which is in the magnetic field (in m) and 

B is the magnetic field strength (in teslas, T). 

The direction of the Lorentz force is perpendicular to both the direction of the flow of current and the 
magnetic field and can be found using the Right Hand Rule as shown in the picture below. Use your right 
hand; your thumb points in the direction of the current, your first finger in the direction of the magnetic 
field and your third finger will then point in the direction of the force. 



Image notjinished 

Figure 13.6 



Both motors and generators can be explained in terms of a coil that rotates in a magnetic field. In a 
generator the coil is attached to an external circuit that is turned, resulting in a changing flux that induces 
an emf. In a motor, a current-carrying coil in a magnetic field experiences a force on both sides of the coil, 
creating a twisting force (called a torque, pronounce like 'talk') which makes it turn. 

Any coil carrying current can feel a force in a magnetic field. The force is the Lorentz force on the moving 
charges in the conductor. The force on opposite sides of the coil will be in opposite directions because the 
charges are moving in opposite directions. This means the coil will rotate, see the picture below: 



247 



Image notjinished 



Figure 13.7 



Instead of rotating the loops through a magnetic field to create electricity, a current is sent through the 
wires, creating electromagnets. The outer magnets will then repel the electromagnets and rotate the shaft 
as an electric motor. If the current is AC, the two slip rings are required to create an AC motor. An AC 
motor is shown in Figure 13.8 



Image notjinished 

Figure 13.8: Layout of an alternating current motor. 



If the current is DC, split-ring commutators are required to create a DC motor. This is shown in 
Figure 13.9. 



Image notjinished 

Figure 13.9: Layout of a direct current motor. 



Image notjinished 

Figure 13.10 



run demo^ 

13.1.2.3 Real-life applications 

13.1.2.3.1 Cars 

A car contains an alternator that charges its battery and powers the car's electric system when its engine 
is running. Alternators have the great advantage over direct-current generators of not using a commutator, 
which makes them simpler, lighter, less costly, and more rugged than a DC generator. 



^http://phet. colorado.edu/sims/faraday/ far adayen.jnlp 



248 CHAPTER 13. ELECTRODYNAMICS 

13.1.2.3.1.1 Research Topic : Alternators 

Try to find out the different ampere values produced by ahernators for different types of machines. Compare 
these to understand what numbers make sense in the real world. You will find different numbers for cars, 
trucks, buses, boats etc. Try to find out what other machines might have alternators. 

A car also contains a DC electric motor, the starter motor, to turn over the engine to start it. A starter 
motor consists of the very powerful DC electric motor and starter solenoid that is attached to the motor. A 
starter motor requires a very high current to crank the engine, that's why it's connected to the battery with 
large cables. 

13.1.2.3.2 Electricity Generation 

AC generators are mainly used in the real-world to generate electricity. 

Image not finished 

Figure 13.11: AC generators are used at the power plant to generate electricity. 



13.1.2.4 Exercise - generators and motors 

1. State the difference between a generator and a motor. 

2. Use Faraday's Law to explain why a current is induced in a coil that is rotated in a magnetic field. 

3. Explain the basic principle of an AC generator in which a coil is mechanically rotated in a magnetic 
field. Draw a diagram to support your answer. 

4. Explain how a DC generator works. Draw a diagram to support your answer. Also, describe how a 
DC generator differs from an AC generator. 

5. Explain why a current-carrying coil placed in a magnetic field (but not parallel to the field) will turn. 
Refer to the force exerted on moving charges by a magnetic field and the torque on the coil. 

6. Explain the basic principle of an electric motor. Draw a diagram to support your answer. 

7. Give examples of the use of AC and DC generators. 

8. Give examples of the uses of motors. 



13.2 Alternating current, inductance and capacitance^ 

13.2.1 Alternating Current 

Most students learning about electricity begin with what is known as direct current (DC), which is electricity 
fiowing in a constant direction. DC is the kind of electricity made by a battery, with definite positive and 
negative terminals. 

However, we have seen that the electricity produced by some generators alternates and is therefore known 
as alternating current (AC). The main advantage to AC is that the voltage can be changed using transformers. 
That means that the voltage can be stepped up at power stations to a very high voltage so that electrical 
energy can be transmitted along power lines at low current and therefore experience low energy loss due to 
heating. The voltage can then be stepped down for use in buildings and street lights. 



^This content is available online at <http://siyavula.cnx.Org/content/m39510/l.l/>. 



249 



NOTE: In South Africa alternating current is generated at a frequency of 50 Hz. 
The circuit symbol for alternating current is: 

Image notjinished 

Figure 13.12 

Graphs of voltage against time and current against time for an AC circuit are shown in Figure 13.13 

Image notjinished 

Figure 13.13: Graph of current or voltage in an AC circuit. 



In an ideal DC circuit the current and voltage are constant. In an AC circuit the current and voltage 
vary with time. The value of the current or voltage at any specific time is called the instantaneous current 
or voltage and is calculated as follows: 

i = ImaxSin{2TTft+ (p) / -, 'i A^ 

(i-oA) 

V = Vmax sin {2tt ft) 

i is the instantaneous current. Imax is the maximum current, v is the instantaneous voltage. Vmax is the 
maximum voltage. / is the frequency of the AC and t is the time at which the instantaneous current or 
voltage is being calculated. 

The average value we use for AC is known as the root mean square (rms) average. This is defined as: 



^rms 



y _ vt^ (13.5) 

* rms — /7^ 

Since AC varies sinusoidally, with as much positive as negative, doing a straight average would get you zero 
for the average voltage. The rms value by-passes this problem. 

13.2.1.1 Exercise - alternating current 

1. Explain the advantages of alternating current. 

2. Write expressions for the current and voltage in an AC circuit. 

3. Define the rms (root mean square) values for current and voltage for AC. 

4. What is the period of the AC generated in South Africa? 

5. If Vmax at a power station generator is 340 V AC, what is the mains supply (rms voltage) in our 
household? 

6. Draw a graph of voltage vs time and current vs time for an AC circuit. 



250 CHAPTER 13. ELECTRODYNAMICS 

13.2,2 Capacitance and inductance 

Capacitors and inductors are found in many circuits. Capacitors store an electric field, and are used as 
temporary power sources as well as to minimize power fiuctuations in major circuits. Inductors work in 
conjunction with capacitors for electrical signal processing. Here we explain the physics and applications of 
both. 

13.2.2.1 Capacitance 

You have learnt about capacitance and capacitors in Grade 11. 

In this section you will learn about capacitance in an AC circuit. A capacitor in an AC circuit has 
reactance. Reactance in an AC circuit plays a similar role to resistance in a DC circuit. The reactance of a 
capacitor Xc is defined as: 

where C is the capacitance and / is the AC frequency. 

If we examine the equation for the reactance of a capacitor, we see that the frequency is in the denomi- 
nator. Therefore, when the frequency is low, the capacitive reactance is very high. This is why a capacitor 
blocks the fiow of DC and low frequency AC because its reactance increases with decreasing frequency. 

When the frequency is high, the capacitive reactance is low. This is why a capacitor allows the fiow of 
high frequency AC because its reactance decreases with increasing frequency. 

13.2.2.2 Inductance 

Inductance (measured in henries, symbol H) is a measure of the generated emf for a unit change in current. 
For example, an inductor with an inductance of 1 H produces an emf of 1 V when the current through the 
inductor changes at the rate of 1 A-s~^. 

An inductor is a passive electrical device used in electrical circuits for its property of inductance. An 
inductor is usually made as a coil (or solenoid) of conducting material, typically copper wire, wrapped around 
a core either of air or of ferromagnetic material. 

Electrical current through the conductor creates a magnetic fiux proportional to the current. A change 
in this current creates a change in magnetic fiux that, in turn, generates an emf that acts to oppose this 
change in current. 

The inductance of an inductor is determined by several factors: 



• 



the shape of the coil; a short, fat coil has a higher inductance than one that is thin and tall. 



• the material that the conductor is wrapped around. 



• 



how the conductor is wound; winding in opposite directions will cancel out the inductance effect, and 
you will have only a resistor. 



The inductance of a solenoid is defined by: 

L = ^ (13.7) 

where /ig is the permeability of the core material (in this case air), A is the cross-sectional area of the 
solenoid, N is the number of turns and I is the length of the solenoid. 

Definition 13.5: Permeability 

Permeability is the property of a material which describes the magnetisation developed in that 
material when excited by a source. 

NOTE: The permeability of free space is Att x 10~^ henry per metre. 



251 

Exercise 13.1: Inductance I (Solution on p. 253.) 

Determine the inductance of a coil with a core material of air. A cross-sectional area of 0, 3m^, 
with 1000 turns and a length of 0,1 m 

Exercise 13.2: Inductance II (Solution on p. 253.) 

Calculate the inductance of a 5 cm long solenoid with a diameter of 4 mm and 2000 turns. 

An inductor in an AC circuit also has a reactance, X^. Reactance is the property of an inductor that opposes 
the flow of AC current. Reactance is defined by: 

Xl = 27r/L (13.8) 

where L is the inductance and / is the frequency of the AC. 

If we examine the equation for the reactance of an inductor, we see that inductive reactance increases 
with increasing frequency. Therefore, when the frequency is low, the inductive reactance is very low. This is 
why an inductor allows the flow of DC and low frequency AC because its reactance decreases with decreasing 
frequency. 

When the frequency is high, the inductive reactance is high. This is why an inductor blocks the flow of 
high frequency AC because its reactance increases with increasing frequency. 

13.2.2.3 Exercise - capacitance and inductance 

1. Describe what is meant by reactance. 

2. Define the reactance of a capacitor. 

3. Explain how a capacitor blocks the flow of DC and low frequency AC but allows the flow of high 
frequency AC. 

4. Describe what is an inductor. 

5. Describe what is inductance. 

6. What is the unit of inductance? 

7. Define the reactance of an inductor. 

8. Write the equation describing the inductance of a solenoid. 

9. Explain how an inductor blocks high frequency AC, but allows low frequency AC and DC to pass. 



13.2,3 Summary 

1. Electrical generators convert mechanical energy into electrical energy. 

2. Electric motors convert electrical energy into mechanical energy. 

3. There are two types of generators - AC and DC. An AC generator is also called an alternator. 

4. There are two types of motors - AC and DC. 

5. Alternating current (AC) has many advantages over direct current (DC). 

6. Capacitors and inductors are important components in an AC circuit. 

7. The reactance of a capacitor or inductor is affected by the frequency of the AC. 



13.2,4 End of chapter exercise 

SC 2003/11 Explain the difference between alternating current (AC) and direct current (DC). 

1. Explain how an AC generator works. You may use sketches to support your answer. 

2. What are the advantages of using an AC motor rather than a DC motor. 

3. Explain how a DC motor works. 

4. At what frequency is AC generated by Eskom in South Africa? 



252 CHAPTER 13. ELECTRODYNAMICS 

5. - Work, Energy and Power in Electric Circuits Mr. Smith read through the agreement with 
Eskom (the electricity provider). He found out that alternating current is supplied to his house at 
a frequency of 50 Hz. He then consulted a book on electric current, and discovered that alternating 
current moves to and fro in the conductor. So he refused to pay his Eskom bill on the grounds that 
every electron that entered his house would leave his house again, so therefore Eskom had supplied him 
with nothing! Was Mr. Smith correct? Or has he misunderstood something about what he is paying 
for? Explain your answer briefly. 

6. What do we mean by the following terms in electrodynamics? 

a. inductance 

b. reactance 

c. solenoid 

d. permeability 



253 



Solutions to Exercises in Chapter 13 

Solution to Exercise 13.1 (p. 251) 

Step 1. We are calculating inductance, so we use the equation: 



Step 2. 



I 
The permeability is that for free space:47ra;10~^ henry per metre 

^ - I 

_ (47rxl0-'')(0,3)(1000) 



L = ^"^^ (13.9) 



(13.10) 



0,1 

= 3,8xlO"^iJ/m 
Step 3. The inductance of the coil is 3, 8 x 10^^ H/m. 
Solution to Exercise 13.2 (p. 251) 

Step 1. Again this is an inductance problem, so we use the same formula as the worked example above. 

4mm 

2mm = 0,002m (13.11) 



2 
A = 7rr2 = 7r X 0,002^ (13.12) 

_ lipAN^ 

I 
_ 47rxlO~'^xO,002^X7rx2000^ 

0,00126i7 
= l,26mH 



254 CHAPTER 13. ELECTRODYNAMICS 



Chapter 14 

Electronics 



14.1 Capacitive and inductive circuits' 

14.1.1 Introduction 

Electronics and electrical devices surround us in daily life. From the street lights and water pumps to 
computers and digital phones, electronics have enabled the digital revolution to occur. All electronics are 
built on a backbone of simple circuits, and so an understanding of circuits is vital in understanding more 
complex devices. 

This chapter will explain the basic physics principles of many of the components of electronic devices. 
We will begin with an explanation of capacitors and inductors. We see how these are used in tuning a radio. 
Next, we look at active components such as transistors and operational amplifiers. Lastly, the chapter will 
finish with an explanation of digital electronics, including logic gates and counting circuits. 

Before studying this chapter, you will want to remind yourself of: 

1. The meaning of voltage {V), current (/) and resistance {R), as covered in Grade 10, and Grade 11. 

2. Capacitors in electric circuits, as covered in Grade 11 (see section 17.6). 

3. Semiconductors, as covered in Grade 11 (see chapter 20). 

4. The meaning of an alternating current (see section 28.3). 

5. Capacitance (C) and Inductance (L) (see section 28.4). 



14.1.2 Capacitive and Inductive Circuits 

Earlier in Grade 12, you were shown alternating currents (AC) and you saw that the voltage and the current 
varied with time. If the AC supply is connected to a resistor, then the current and voltage will be proportional 
to each other. This means that the current and voltage will 'peak' at the same time. We say that the current 
and voltage are in phase. This is shown in Figure 14.1. 

Image notjinished 

Figure 14.1: The voltage and current are in phase when a resistor is connected to an alternating voltage. 



^This content is available online at <http://siyavula.cnx.Org/content/m39523/l.l/>. 

255 



256 CHAPTER 14. ELECTRONICS 

When a capacitor is connected to an alternating voltage, the maximum voltage is proportional to the 
maximum current, but the maximum voltage does not occur at the same time as the maximum current. The 
current has its maximum (it peaks) one quarter of a cycle before the voltage peaks. Engineers say that the 
'current leads the voltage by 90°'. This is shown in Figure 14.2. 



Image notjinished 



Figure 14.2: The current peaks (has its maximum) one quarter of a wave before the voltage when a 
capacitor is connected to an alternating voltage. 



For a circuit with a capacitor, the instantaneous value of y is not constant. However, the value of y"""" is 
useful, and is called the capacitive reactance (Xc) of the component. Because it is still a voltage divided 
by a current (like resistance), its unit is the ohm. The value of Xc (C standing for capacitor) depends on 
its capacitance (C) and the frequency (/) of the alternating current (in South Africa 50 Hz). 

Inductors are very similar, but the current peaks 90° after the voltage. This is shown in Figure 14.3. 
Engineers say that the 'current lags the voltage'. Again, the ratio of maximum voltage to maximum current 
is called the reactance — this time inductive reactance (Xj^). The value of the reactance depends on its 
inductance (L). 

Xl = ^ = 2TTfL 



Image notjinished 



Figure 14.3: The current peaks (has its maximum) one quarter of a wave after the voltage when an 
inductor is connected to an alternating voltage. 



Definition 14.1: Reactance 

The ratio of the maximum voltage to the maximum current when a capacitor or inductor is 
connected to an alternating voltage. The unit of reactance is the ohm. 

While inductive and capacitive reactances are similar, in one sense they are opposites. For an inductor, 
the current peaks 90° after the voltage. For a capacitor the current peaks 90°ahead of the voltage. When we 
work out the total reactance for an inductor and a capacitor in series, we use the formula Xtotai = Xl — Xc 

to take this into account. This formula can also be used when there is more than one inductor or more 
than one capacitor in the circuit. The total reactance is the sum of all of the inductive reactances minus the 
sum of all the capacitive reactances. The magnitude (number) in the final result gives the ratio of maximum 
voltage to maximum current in the circuit as a whole. The sign of the final result tells you its phase. If it is 
positive, the current peaks 90° after the voltage, if it is negative, the current peaks 90° before the voltage. 

If a series circuit contains resistors as well, then the situation is more complicated. The maximum current 
is still proportional to the maximum voltage, but the phase difference between them won't be 90°. The ratio 
between the maximum voltage and maximum current is called the impedance {Z), and its unit is also the 
ohm. Impedances are calculated using this formula: Z = yJX"^ + B? 



257 

where X is the total reactance of the inductors and capacitors in the circuit, and R is the total resistance 
of the resistors in the circuit. 

It is easier to understand this formula by thinking of a right angled triangle. Resistances are drawn 
horizontally, reactances are drawn vertically. The hypotenuse of the triangle gives the impedance. This is 
shown in Figure 14.4. 

Image notjtnished 

Figure 14.4: Visualizing the relationship between reactance, resistance and impedance. 



Definition 14.2: Impedance 

The maximum voltage divided by the maximum current for any circuit. The unit of impedance is 
the ohm. 

It is important to remember that when resistors and inductances (or capacitors) are in a circuit, the 
current will not be in phase with the voltage, so the impedance is not a resistance. Similarly the current 
won't be exactly 90° out of phase with the voltage so the impedance isn't a reactance either. 

This simulation allows you to build some circuits with capacitors and inductors. 



Image notjinished 

Figure 14.5 



run demo^ 

Exercise 14.1: The impedance of a coil (Solution on p. 281.) 

Calculate the maximum current in a coil in a South African motor which has a resistance of 5 il 
and an inductance of 3 mH. The maximum voltage across the coil is 6 V. You can assume that the 
resistance and inductance are in series. 

Exercise 14.2: An RC circuit (Solution on p. 281.) 

Part of a radio contains a 30 i7 resistor in series with a 3 /xF capacitor. What is its impedance at 
a frequency of 1 kHz? 



14.1.2.1 Capacitive and Inductive Circuits 

1. Why is the instantaneous value of y of little use in an AC circuit containing an inductor or capacitor? 

2. How is the reactance of an inductor different to the reactance of a capacitor? 

3. Why can the ratio of the maximum voltage to the maximum current in a circuit with a resistor and 
an inductor not be called a reactance? 



^http://phet. Colorado. edu/sims/circuit-construction-kit/circuit-construction-kit-ac-virtual-laben.jnlp 



258 CHAPTER 14. ELECTRONICS 

4. An engineer can describe a motor as equivalent to a 30 i7 resistor in series with a 30 niH inductor. If 
the maximum value of the supply voltage is 350 V, what is the maximum current? Assume that the 
frequency is 50 Hz. 

5. A timer circuit in a factory contains a 200 /iF capacitor in series with a 10 kil resistor. What is its 
impedance? Assume that the frequency is 50 Hz. 

6. A 3 mH inductor is connected in series with a 100 /xF capacitor. The reactance of the combination is 
zero. What is the frequency of the alternating current? 

NOTE: Most factories containing heavy duty electrical equipment (e.g. large motors) have to pay 
extra money to their electricity supply company because the inductance of the motor coils causes 
the current and voltage to get out of phase. As this makes the electricity distribution network 
less efficient, a financial penalty is incurred. The factory engineer can prevent this by connecting 
capacitors into the circuit to reduce the reactance to zero, as in the last question above. The current 
and voltage are then in phase again. We can't calculate the capacitance needed in this chapter, 
because the capacitors are usually connected in parallel, and we have only covered the reactances 
and impedances of series circuits. 



14.1.3 Filters and Signal Tuning 

14.1.3.1 Capacitors and Inductors as Filters 

We have already seen how capacitors and inductors pass current more easily at certain frequencies than 
others. To recap: if we examine the equation for the reactance of a capacitor, we see that the frequency 
is in the denominator. Therefore, when the frequency is low, the capacitive reactance is very high. This is 
why a capacitor blocks the flow of DC and low frequency AC because its reactance increases with decreasing 
frequency. 

When the frequency is high, the capacitive reactance is low. This is why a capacitor allows the flow of 
high frequency AC because its reactance decreases with increasing frequency. 

Therefore putting a capacitor in a circuit blocks the low frequencies and allows the high frequencies to 
pass. This is called a high pass filter. A filter like this can be used in the 'treble' setting of a sound mixer 
or music player which controls the amount of high frequency signal reaching the speaker. The more high 
frequency signal there is, the 'tinnier' the sound. A simple high pass filter circuit is shown in Figure 14.6. 



Image not finished 



Figure 14.6: A high pass filter. High frequencies easily pass through the capacitor and into the next 
part of the circuit, while low frequencies pass through the inductor straight to ground. 



Similarly, if we examine the equation for the reactance of an inductor, we see that inductive reactance 
increases with increasing frequency. Therefore, when the frequency is low, the inductive reactance is very 
low. This is why an inductor allows the flow of DC and low frequency AC because its reactance decreases 
with decreasing frequency. 

When the frequency is high, the inductive reactance is high. This is why an inductor blocks the flow of 
high frequency AC because its reactance increases with increasing frequency. 

Therefore putting an inductor in a circuit blocks the high frequencies and allows the low frequencies to 
pass. This is called a low pass filter. A filter like this can be used in the 'bass' setting of a sound mixer or 



259 

music player which controls the amount of low frequency signal reaching the speaker. The more low frequency 
signal there is, the more the sound 'booms'. A simple low pass filter circuit is shown in Figure 14.7. 



Image notjtnished 



Figure 14.7: A low pass filter. Low frequencies pass through the inductor and into the next part of the 
circuit, while high frequencies pass through the capacitor straight to ground. 



14.1.3.2 LRC Circuits, Resonance and Signal Tuning 

A circuit containing a resistor, a capacitor and an inductor all in series is called an LRC circuit. Because the 
components are in series, the current through each component at a particular time will be the same as the 
current through the others. The voltage across the resistor will be in phase with the current. The voltage 
across the inductor will be 90° ahead of the current (the current always follows or lags the voltage in an 
inductor). The voltage across the capacitor will be 90° behind the current (the current leads the voltage for 
a capacitor) . The phases of the voltages for the inductor and capacitor relative to the common current are 
shown in Figure 14.8. 



Image notjinished 



Figure 14.8: The voltages across the separate components of an LRC circuit. Looking at the peaks, 
you see that the voltage across the inductor Vl 'peaks' first, followed 90° later by the current /, followed 
90° later by the voltage across the capacitor Vc- The voltage across the resistor is not shown — it is in 
phase with the current and peaks at the same time as the current. 



The reactance of the inductor is 27r/L, and the reactance of the capacitor is l/27r/C but with the 
opposite phase. So the total reactance of the LRC circuit is X = Xl — Xq = ^nfL — 277c '^^^ impedance 



of the circuit as a whole is given by Z = \J X'^ + R"^ = \ ( 27r/L — 2i7c ) + -R^ At different frequencies, the 

impedance will take different values. The impedance will have its smallest value when the positive inductive 
reactance cancels out the negative capacitive reactance. This occurs when 27r/L = ^ \^g so the frequency 
of minimum impedance must be / = - — ^==^ This is called the resonant frequency of the circuit. This is 
the frequency at which you can get the largest current for a particular supply voltage. It is also called the 
natural frequency of the circuit. This means the frequency at which the circuit would oscillate by itself. 

Definition 14.3: Resonance 

Resonance occurs when a circuit is connected to an alternating voltage at its natural frequency. A 
very large current can build up in the circuit, even with minimal power input. 

An LRC circuit is very useful when we have a signal containing many different frequencies, and we only 
want one of them. If a mixed signal like this is connected to an LRC circuit, then only the resonant frequency 
(and other frequencies close to it) will drive measureable currents. This means that an LRC circuit can select 
one frequency from a range. This process is called signal tuning. 



260 CHAPTER 14. ELECTRONICS 

NOTE: When you set up a radio antenna, and amplify the radio signal it receives, you find many 
different bands of frequencies — one from each radio station. When you listen to the radio, you 
only want to listen to one station. An LRC circuit in the radio (the tuning circuit) is set so that 
its resonant frequency is the frequency of the station you want to listen to. This means that of 
the many radio stations broadcasting, you only hear one. When you adjust the tuning dial on the 
radio, it changes the capacitance of the capacitor in the tuning circuit. This changes the resonant 
frequency, and this changes the radio station that you will hear. 



14.1.3.2.1 Filters and Signal Tuning 

1. Which component would you use if you wanted to block low frequencies? 

2. Which component would you use if you wanted to block high frequencies? 

3. Calculate the impedance of a series circuit containing a 50 fi resistor, a 30 /iF capacitor and a 3 mH 
inductor for frequencies of (a) 50 Hz, (b) 500 Hz, and (c) 5 000 Hz. 

4. Calculate the resonant frequency of the circuit in the previous question, part (c). 

5. A radio station broadcasts at a frequency of 150 kHz. The tuning circuit in the radio contains a 0.3 
mH inductor. What is the capacitance of the capacitor needed in the tuning circuit if you want to 
listen to this radio station? 

6. State the relationship between the phase of the voltages across an inductor, a resistor and a capacitor 
in an LRC circuit. 

7. Explain what is meant by resonance. 

8. Explain how LRC circuits are used for signal tuning, for example in a radio. 



14.2 Principles of digital electronics^ 

14.2.1 The Principles of Digital Electronics 

The circuits and components we have discussed are very useful. You can build a radio or television with 
them. You can make a telephone. Even if that was all there was to electronics, it would still be very useful. 
However, the great breakthrough in the last fifty years or so has been in digital electronics. This is the 
subject which gave us the computer. The computer has revolutionized the way business, engineering and 
science are done. Small computers programmed to do a specific job (called microprocessors) are now used 
in almost every electronic machine from cars to washing machines. Computers have also changed the way 
we communicate. We used to have telegraph or telephone wires passing up and down a country — each one 
carrying one telephone call or signal. We now have optic fibres each capable of carrying tens of thousands 
of telephone calls using digital signals. 

So, what is a digital signal? Look at Figure 14.9. A normal signal, called an analogue signal, carries a 
smooth wave. At any time, the voltage of the signal could take any value. It could be 2,00 V or 3,53 V or 
anything else. A digital signal can only take certain voltages. The simplest case is shown in the figure — 
the voltage takes one of two values. It is either high, or it is low. It never has any other value. 

These two special voltages are given symbols. The low voltage level is written 0, while the high voltage 
level is written as 1. When you send a digital signal, you set the voltage you want (0 or 1), then keep this 
fixed for a fixed amount of time (for example 0.01 ^s), then you send the next or 1. The digital signal in 
Figure 14.9 could be written 01100101. 



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261 

Image notjinished 

Figure 14.9: The difference between normal (analogue) signals and digital signals. 

Why are digital signals so good? 

1. Using a computer, any information can be turned into a pattern of Os and Is. Pictures, recorded music, 
text and motion pictures can all be turned into a string of Os and Is and transmitted or stored in the 
same way. The computer receiving the signal at the other end converts it back again. A Compact Disc 
(CD) for example, can store music or text or pictures, and all can be read using a computer. 

2. The and the 1 look very different. You can immediately tell if a or a 1 is being sent. Even if there 
is interference, you can still tell whether the sender sent a or a 1. This means that fewer mistakes 
are made when reading a digital signal. This is why the best music recording technologies, and the 
most modern cameras, for example, all use digital technology. 

3. Using the Os and Is you can count, and do all kinds of mathematics. This will be explained in more 
detail in the next section. 

The simplest digital circuits are called logic gates. Each logic gate makes a decision based on the information 
it receives. Different logic gates are set up to make the decisions in different ways. Each logic gate will be 
made of many microscopic transistors connected together within a thin wafer of silicon. This tiny circuit is 
called an Integrated Circuit or I.C. - all the parts are in one place (integrated) on the silicon wafer. 

14.2.1.1 Logic Gates 

There are five main types of logic gate: NOT, AND, OR, NAND and NOR. Each one makes its decision in 
a different way. 

14.2.1.1.1 The NOT Gate 

Problem: You want an automatic circuit in your ofRce to turn on the heating in the winter. You already 
have a digital electronic temperature sensor. When the temperature is high, it sends out a 1. When the 
ofRce is cold, it sends out a 0. If this signal were sent straight to the heater, the heater would turn on (1) 
when it was already hot, and would stay off when it was cold. This is wrong! To make the heater work, 
we need a circuit which will change a (from the sensor) into a 1 (to send to the heater). This will make 
the heater come on when it is cold. You also want it to change a 1 (from the sensor) into a (to send to 
the heater). This will turn the heater off when the room is hot. This circuit is called an inverter or NOT 
gate. It changes into 1 (1 is NOT 0). It changes 1 into (0 is NOT 1). It changes a signal into what it is 
NOT. 

The symbol for the NOT gate is: 



Image notjinished 

Figure 14.10 



262 



CHAPTER 14. ELECTRONICS 



The action of the NOT gate can be written in a table called a truth table. The left column shows the 
possible inputs on different rows. The right column shows what the output (decision) of the circuit will be 
for that input. The truth table for the NOT gate is shown below. 



Input 


Output 





1 


1 






Table 14.1 

When you read the truth table, the top row says, "If the input is 0, the output will be 1." For our heater, 
this means, "If the room is cold, the heater will turn on." The bottom row says, "If the input is 1, the output 
will be 0." For our heater, this means, "If the room is hot, the heater will switch off." 

14.2.1.1.2 The AND Gate 

Problem: An airliner has two toilets. Passengers get annoyed if they get up from their seat only to find that 
both toilets are being used and they have to go back to their seat and wait. You want to fit an automatic 
circuit to light up a display if both toilets are in use. Then passengers know that if the light is off, there will 
be a free toilet for them to use. There is a sensor in each toilet. It gives out a of the toilet is free, and a 1 
if it is in use. You want to send a 1 to the display unit if both sensors are sending Is. To do this, you use 
an AND gate. 

The symbol for the AND gate is: 



Image notjinished 



Figure 14.11: Symbol for the AND logic gate. 



The truth table for the AND gate is shown below. An AND gate has two inputs (the NOT gate only had 
one). This means we need four rows in the truth table, one for each possible set of inputs. The first row, 
for example, tells us what the AND gate will do if both inputs are 0. In our airliner, this means that both 
toilets are free. The right column has a showing that the output will be 0, so the display will not light 
up. The second row has inputs of and 1 (the first toilet is free, the other is in use). Again the output is 0. 
The third row tells us what will happen if the inputs are 1 and (the first toilet is in use, and the second 
is free). Finally, the last line tells us what will happen if both inputs are 1 (the first toilet is in use and the 
second toilet is in use). It is only in this case that the output is 1 and the display lights up. 



Inputs 


Output 


A 


B 
















1 





1 








1 


1 


1 



263 



Table 14.2 

This device is called an AND gate, because the output is only 1 if one input AND the other input are 
both 1. 

14.2.1.1.2.1 Using and 1 to mean True and False 

When we use logic gates we use the low voltage state to represent 'false'. The high voltage state 1 represents 
'true'. This is why the word AND is so appropriate. A AND B is true (1) if, and only if, A is true (1) AND 
B is true (1). 

14.2.1.1.2.2 AND and multiplication 

Sometimes, the AND operation is written as multiplication. A AND B is written AB. If either A or B are 
0, then AB will also be 0. For AB to be 1, we need A and B to both be 1. Multiplication of the numbers 
and 1 does exactly the same job as an AND gate. 

14.2.1.1.3 The NAND Gate 

Problem: You build the circuit for the airliner toilets using an AND gate. Your customer is pleased, but 
she says that it would be better if the display lit up when there was a free toilet. In other words, the display 
should light up unless both toilets are in use. To do this we want a circuit which does the opposite of an 
AND gate. We want a circuit which would give the output where an AND gate would give 1. We want a 
circuit which would give the output 1 where an AND gate would give 0. This circuit is called a NAND gate. 
The symbol for the NAND gate is: 



Image notjinished 



Figure 14.12 



The truth table for the NAND gate is shown below. 



Inputs 


Output 


A 


B 










1 





1 


1 


1 





1 


1 


1 






Table 14.3 



You may have noticed that we could have done this job on the airliner by using our earlier circuit, with 
a NOT gate added between the original AND gate and the display. This is where the word NAND comes 
from — it is short for Not AND. 



264 



CHAPTER 14. ELECTRONICS 



14.2.1.1.4 The OR Gate 

Problem: A long, dark corridor has two Hght switches — one at each end of the corridor. The switches 
each send an output of to the control unit if no-one has pressed the switch. If someone presses the switch, 
its output is 1. The lights in the corridor should come on if either switch is pressed. To do this job, the 
control unit needs an OR gate. The symbol for the OR gate is: 



Image notjinished 



Figure 14.13: Symbol for the OR logic gate. 



The truth table for the OR gate is shown. 



Inputs 


Output 


A 


B 
















1 


1 


1 





1 


1 


1 


1 



Table 14.4 

You can see that the output is 1 (and the lights come on in the corridor) if either one switch OR the 
other is pressed. Pressing both switches also turns on the lights, as the last row in the table shows. 

14.2.1.1.4.1 OR and addition 

Sometimes you will see A OR B written mathematically as A+B. This makes sense, since if A=0 and B=0, 
then A OR B = A+B = 0. Similarly, if A=0 and B=l, then A OR B = A+B = 1. If A=l and B=0, then A 
OR B = A+B = 1 once again. The only case where the OR function differs from normal addition is when 
A=l and B=l. Here A OR B = 1 in logic, but A+B=2 in arithmetic. However, there is no such thing as 
'2' in logic, so we define + to mean 'OR', and write 1+1=1 with impunity! 

If you wish, you can prove that the normal rules of algebra still work using this notation: A+(B+C) = 
(A+B)+C, A(BC) = (AB)C, and A(B+C) = AB + AC. This special kind of algebra where variables can 
only be (representing false) or 1 (representing true) is called Boolean algebra. 



14.2.1.1.5 The NOR Gate 

The last gate you need to know is the NOR gate. This is opposite to the OR gate. The output is 1 if both 
inputs are 0. In other words, the output switches on if neither the first NOR the second input is 1. The 
symbol for the NOR gate is: 



265 



Image notjinished 

Figure 14.14: Symbol for the NOR logic gate. 
The truth table for the NOR gate is shown below. 



Inputs 


Output 


A 


B 










1 





1 





1 








1 


1 






Table 14.5 

The examples given were easy. Each job only needed one logic gate. However any 'decision making' 
circuit can be built with logic gates, no matter how complicated the decision. Here is an example. 

Exercise 14.3: An Economic Heating Control (Solution on p. 281.) 

A sensor in a building detects whether a room is being used. If it is empty, the output is 0, if it 
is in use, the output is 1. Another sensor measures the temperature of the room. If it is cold, the 
output is 0. If it is hot, the output is 1. The heating comes on if it receives a 1. Design a control 
circuit so that the heating only comes on if the room is in use and it is cold. 

Exercise 14.4: Solving a circuit with two logic gates (Solution on p. 281.) 

Compile the truth table for the circuit below. 



Image notjinished 



Figure 14.15 



Each logic gate is manufactured from two or more transistors. Other circuits can be made using logic gates, 
as we shall see in the next section. We shall show you how to count and store numbers using logic gates. 
This means that if you have enough transistors, and you connect them correctly to make the right logic 
gates, you can make circuits which count and store numbers. 

In practice, the cheapest gate to manufacture is usually the NAND gate. Additionally, Charles Peirce 
showed that NAND gates alone (as well as NOR gates alone) can be used to reproduce all the other logic 
gates. 

14.2.1.1.5.1 The Principles of Digital Electronics 

1. Why is digital electronics important to modern technology and information processing? 



266 CHAPTER 14. ELECTRONICS 

2. What two symbols are used in digital electronics, to represent a "high" and a "low"? What is this 
system known as? 

3. What is a logic gate? 

4. What are the five main types of logic gates? Draw the symbol for each logic gate. 

5. Write out the truth tables for each of the five logic gates. 

6. Write out the truth table for the following circuit. Which single gate is this circuit equivalent to? 

Image notjtnished 

Figure 14.16 

7. Write out the truth table for the following circuit. Which single gate is this circuit equivalent to? 

Image notjtnished 

Figure 14.17 



14.2.2 Using and Storing Binary Numbers 

In the previous section, we saw how the numbers and 1 could represent 'false' and 'true' and could be used 
in decision making. Often we want to program a computer to count with numbers. To do this we need a 
way of writing any number using nothing other than and 1. When written in this way, numbers are called 
binary numbers. 

Definition 14.4: Binary Number System 

A way of writing any number using only the digits and 1. 

14.2.2.1 Binary numbers 

In normal (denary) numbers, we write 9+1 as 10. The fact that the '1' in 10 is the second digit from the 
right tells us that it actually means 10 and not 1. Similarly, the '3' in 365 represents 300 because it is the 
third digit from the right. You could write 365 as 3 x 100 + 6 x 10 + 5. You will notice the pattern that the 
nth digit from the right represents 10"~^. In binary, we use the nth digit from the right to represent 2"~^. 
Thus 2 is written as 10 in binary. Similarly 2^ = 4 is written as 100 in binary, and 2'^ = 8 is written as 1000 
in binary. 

Exercise 14.5: Conversion of Binary Numbers to Denary Numbers (Solution on p. 

282.) 
Convert the binary number 10101 to its denary equivalent. 

Exercise 14.6: Conversion of Denary Numbers to Binary Numbers (Solution on p. 

282.) 
Convert the decimal number 12 to its binary equivalent. 



267 

NOTE: How do you write numbers as a sum of powers of two? The first power of two (the largest) 
is the largest power of two which is not larger than the number you are working with. In our last 
example, where we wanted to know what twelve was in binary, the largest power of two which is 
not larger than 12 is 8. Thus 12 = 8 + something. By arithmetic, the 'something' must be 4, and 
the largest power of two not larger than this is 4 exactly. Thus 12 = 8 + 4, and we have finished. 

A more complicated example would be to write one hundred in binary. The largest power of two 
not larger than 100 is 64 (1000000 in binary). Subtracting 64 from 100 leaves 36. The largest power 
of two not larger than 36 is 32 (100000 in binary). Removing this leaves a remainder of 4, which 
is a power of two itself (100 in binary). Thus one hundred is 64 + 32 + 4, or in binary 1000000 + 
100000 + 100 = 1100100. 

Once a number is written in binary, it can be represented using the low and high voltage levels of digital 
electronics. We demonstrate how this is done by showing you how an electronic counter works. 

14.2.2.2 Counting circuits 

To make a counter you need several 'T flip flops', sometimes called 'divide by two' circuits. A T flip flop is 
a digital circuit which swaps its output (from to 1 or from 1 to 0) whenever the input changes from 1 to 0. 
When the input changes from to 1 it doesn't change its output. It is called a flip flop because it changes 
(flips or flops) each time it receives a pulse. 

If you put a series of pulses 10101010 into a T flip flop, the result is 01100110. Figure 14.18 makes this 
clearer. 



Image notjinished 



Figure 14.18: The output of a T flip flop, or 'divide by two' circuit when a square wave is connected 
to the input. The output changes state when the input goes from 1 to 0. 



As you can see from Figure 14.18, there are half as many pulses in the output. This is why it is called a 
'divide by two' circuit. 

If we connect T flip flops in a chain, then we make a counter which can count pulses. As an example, we 
connect three T flip flops in a chain. This is shown in Figure 14.19. 



Image notjinished 



Figure 14.19: Three T flip flops connected together in a chain to make a counter. The input of each 
flip flop is labelled T, while each output is labelled Q. The pulses are connected to the input on the left. 
The outputs Qo, Q\ and Q2 give the three digits of the binary number as the pulses are counted. This 
is explained in the text and in the next table. 



When this circuit is fed with a stream of pulses, the outputs of the different stages change. The table 
below shows how this happens. Each row shows a different stage, with the first stage at the top. We assume 
that all of the flip flops have as their output to start with. 



268 



CHAPTER 14. ELECTRONICS 



Input 


Output 1 


Output 2 


Output 3 


Number of pulse 


Number in binary 


1 














000 













1 


001 


1 










1 


001 








1 





2 


010 


1 





1 





2 


010 







1 





3 


Oil 


1 




1 





3 


Oil 













4 


100 


1 










4 


100 












5 


101 


1 









5 


101 








1 




6 


110 


1 





1 




6 


110 







1 




7 


111 


1 




1 




7 


111 














8 


1000 


1 











8 


1000 













9 


1101 


1 










9 


1101 



Table 14.6 

The binary numbers in the right hand column count the pulses arriving at the input. You will notice 
that the output of the first flip flop gives the right most digit of the pulse count (in binary). The output of 
the second flip flop gives the second digit from the right (the 'twos' digit) of the pulse count. The output 
of the third flip flop gives the third digit from the right (the 'fours' digit) of the pulse count. As there are 
only three flip flops, there is nothing to provide the next digit (the 'eights' digit), and so the eighth pulse is 
recorded as 000, not 1000. 

This device is called a modulo 8 counter because it can count in eight stages from 000 to 111 before it 
goes back to 000. If you put four flip flops in the counter, it will count in sixteen stages from 0000 to 1111, 
and it is called a modulo 16 counter because it counts in sixteen stages before going back to 0000. 

Definition 14.5: Modulo 

The modulo of a counter tells you how many stages (or pulses) it receives before going back to 
as its output. Thus a modulo 8 counter counts in eight stages 000, 001, 010, Oil, 100, 101, 110, 
111, then returns to 000 again. 



NOTE: If a counter contains n flip flops, it will be a modulo 2" counter. It will count from to 

2"- 1. 



269 

14.2.2.3 Storing binary numbers 

Counting is important. However, it is equally important to be able to remember the numbers. Computers 
can convert almost anything to a string of Os and Is, and therefore to a binary number. Unless this number 
can be stored in the computer's memory, the computer would be useless. 

The memory in the computer contains many parts. Each part is able to store a single or 1. Since 
and 1 are the two binary digits, we say that each part of the memory stores one bit. 

Definition 14.6: Bit 

One bit is a short way of saying one 'binary digit'. It is a single or 1. 

NOTE: If you have eight bits, you can store a binary number from 00000000 to 11111111 (0 to 
255 in denary). This gives you enough permutations of Os and Is to have one for each letter of 
the alphabet (in upper and lower case), each digit from to 9, each punctuation mark and each 
control code used by a computer in storing a document. When you type text into a word processor, 
each character is stored as a set of eight bits. Each set of eight bits is called a byte. Computer 
memories are graded according to how many bytes they store. There are 1024 bytes in a kilobyte 
(kB), 1024 X 1024 bytes in a megabyte (MB), and 1024 x 1024 x 1024 bytes in a gigabyte (GB). 

To store a bit we need a circuit which can 'remember' a or a 1. This is called a bistable circuit because it 
has two stable states. It can stay indefinitely either as a or a 1. An example of a bistable circuit is shown 
in . It is made from two NOR gates. 

Image notjinished 

Figure 14.20 

Image notjinished 

Figure 14.21 



A bistable circuit made from two NOR gates. This circuit is able to store one bit of digital information. 
With the two inputs set to 0, you can see that the output could be (and will remain) either or 1. The 
circuit on the top shows an output of 0, the circuit underneath shows an output of 1. Wires carrying high 
logic levels (1) are drawn thicker. The output of the bistable is labelled Q. 

To store the or the 1 in the bistable circuit, you set one of the inputs to 1, then put it back to again. 
If the input labelled 'S' (set) is raised, the output will immediately become 1. This is shown in Figure 14.22. 



270 CHAPTER 14. ELECTRONICS 



Image notjinished 



Figure 14.22: The output of a bistable circuit is set (made 1) by raising the 'S' input to 1. Wires 
carrying high logic levels (1) are shown with thicker lines. 



To store a 0, you raise the 'R' (reset) input to 1. This is shown in Figure 14.23. 



Image notjinished 



Figure 14.23: The output of a bistable circuit is reset (made 0) by raising the 'R' input to 1. Wires 
carrying high logic levels (1) are shown with thicker lines. 



Once you have used the S or R inputs to set or reset the bistable circuit, you then bring both inputs back 
to 0. The bistable 'remembers' the state. Because of the ease with which the circuit can be Reset and Set 
it is also called an RS flip flop circuit. 

Computer memory can store millions or billions of bits. If it used our circuit above, it would need millions 
or billions of NOR gates, each of which is made from several transistors. Computer memory is made of many 
millions of transistors. 

NOTE: The bistable circuits drawn here don't remember Os or Is forever — they lose the information 
if the power is turned off. The same is true for the RAM (Random Access Memory) used to store 
working and temporary data in a computer. Some modern circuits contain special memory which 
can remember its state even if the power is turned off. This is used in FLASH drives, commonly 
found in USB data sticks and on the memory cards used with digital cameras. These bistable 
circuits are much more complex. 

You can also make T flip flops out of logic gates, however these are more complicated to design. 

14.2.2.3.1 Counting Circuits 

1. What is the term bit short for? 

2. What is 43 in binary? 

3. What is 1100101 in denary? 

4. What is the highest number a modulo 64 counter can count to? How many T flip flops does it contain? 

5. What is the difference between an RS flip flop and a T flip flop? 

6. Draw a circuit diagram for a bistable circuit (RS flip flop). Make three extra copies of your diagram. 
On the first diagram, colour in the wires which will carry high voltage levels (digital 1) if the R input 
is low, and the S input is high. On the second diagram, colour in the wires which carry high voltage 
levels if the S input of the first circuit is now made low. On the third diagram, colour in the wires 
which carry high voltage levels if the R input is now made high. On the final diagram, colour in the 
wires carrying high voltage levels if the R input is now made low again. 

7. Justify the statement: a modern computer contains millions of transistors. 



271 

14.2.2.3.2 End of Chapter Exercises 

1. Calculate the reactance of a 3 mH inductor at a frequency of 50 Hz. 

2. Calculate the reactance of a 30 /xF capacitor at a frequency of 1 kHz. 

3. Calculate the impedance of a series circuit containing a 5 niH inductor, a 400 /iF capacitor and a 2 kil 
resistor at a frequency of 50 kHz. 

4. Calculate the frequency at which the impedance of the circuit in the previous question will be the 
smallest. 

5. Which component can be used to block low frequencies? 

6. Draw a circuit diagram with a battery, diode and resistor in series. Make sure that the diode is forward 
biased so that a current will flow through it. 

7. When building a complex electronic circuit which is going to be powered by a battery, it is always a 
good idea to put a diode in series with the battery. Explain how this will protect the circuit if the user 
puts the battery in the wrong way round. 

8. Summarize the differences betwen a bipolar and fleld effect transistor. 

9. What does an operational amplifler (op-amp) do? 

10. What is the difference between a digital signal and an analogue signal? 

11. What are the advantages of digital signals over analogue signals? 

12. Draw the symbols for the flve logic gates, and write down their truth tables. 

13. Draw a circuit diagram with an AND gate. Each input should be connected to the output of a separate 
NOT gate. By writing truth tables show that this whole circuit behaves as a NOR gate. 

14. Convert the denary number 99 into binary. 

15. Convert the binary number 11100111 into denary. 

16. Explain how three T flip flops can be connected together to make a modulo 8 counter. What is the 
highest number it can count up to? 

17. Draw the circuit diagram for an RS flip flop (bistable) using two NOR gates. 

18. Show how the circuit you have just drawn can have a stable output of or 1 when both inputs are 0. 

19. Operational (and other) ampliflers, logic gates, and flip flops all contain transistors, and would not 
work without them. Write a short newspaper article for an intelligent reader who knows nothing about 
electronics. Explain how important transistors are in modern society. 



14.3 Active circuit elements^ 
14.3.1 Active Circuit Elements 

The components you have been learning about so far — resistors, capacitors and inductors — are called 
passive components. They do not change their behaviour or physics and therefore always have the same 
response to changes in voltage or current. Active components are quite different. Their response to changes 
in input allows them to make ampliflers, calculators and computers. 

14.3.1.1 The Diode 

A diode is an electronic device that allows current to flow in one direction only. 



This content is available online at <http://siyavula.cnx.Org/content/m39531/l.l/>. 



272 CHAPTER 14. ELECTRONICS 



Image notjtnished 

Figure 14.24: Diode circuit symbol and direction of flow of current. 



A diode consists of two doped semi-conductors joined together so that the resistance is low when connected 
one way and very high the other way. 



Image notjtnished 



Figure 14.25: Operation of a diode. (Left) The diode is forward biased and current is permitted. The 
negative terminal of the battery is connected to the negative terminal of the diode. (Right) The diode is 
reverse biased and current flow is not allowed. The negative terminal of the battery is connected to the 
positive terminal of the diode. 



A full explanation of diode operation is complex. Here is a simplified description. The diode consists of 
two semiconductor blocks attached together. Neither block is made of pure silicon — they are both doped. 
Doping was described in more detail in Section 10.3. 

In short, p-type semiconductor has fewer free electrons than normal semiconductor. 'P' stands for 
'positive', meaning a lack of electrons, although the material is actually neutral. The locations where 
electrons are missing are called holes. This material can conduct electricity well, because electrons can 
move into the holes, making a new hole somewhere else in the material. Any extra electrons introduced into 
this region by the circuit will fill some or all of the holes. 

In n-type semiconductor, the situation is reversed. The material has more free electrons than normal 
semiconductor. 'N' stands for 'negative', meaning an excess of electrons, although the material is actually 
neutral. 

When a p-type semiconductor is attached to an n-type semiconductor, some of the free electrons in the 
n-type move across to the p-type semiconductor. They fill the available holes near the junction. This means 
that the region of the n-type semiconductor nearest the junction has no free electrons (they've moved across 
to fill the holes). This makes this n-type semiconductor positively charged. It used to be electrically neutral, 
but has since lost electrons. 

The region of p-type semiconductor nearest the junction now has no holes (they've been filled in by the 
migrating electrons). This makes the p-type semiconductor negatively charged. It used to be electrically 
neutral, but has since gained electrons. 

Without free electrons or holes, the central region can not conduct electricity well. It has a high resistance, 
and is called the depletion band. This is shown in Figure 14.26. 



273 



Image notjtnished 



Figure 14.26: A diode consists of two doped semi-conductors joined together so that the resistance is 
low when connected one way and very high the other way. 



You can explain the high resistance in a different way. A free electron in the n-type semiconductor will 
be repelled from the p-type semiconductor because of its negative charge. The electron will not go into the 
depletion band, and certainly won't cross the band to the p-type semiconductor. You may ask, "But won't 
a free electron in the p-type semiconductor be attracted across the band, carrying a current?" But there are 
no free electrons in p-type semiconductor, so no current of this kind can flow. 

If the diode is reverse-biased, the -I- terminal of the battery is connected to the n-type semiconductor. 
This makes it even more negatively charged. It also removes even more of the free electrons near the depletion 
band. At the same time, the — terminal of the battery is connected to the p-type silicon. This will supply free 
electrons and fill in more of the holes next to the depletion band. Both processes cause the depletion band 
to get wider. The resistance of the diode (which was already high) increases. This is why a reverse-biased 
diode does not conduct. 

Another explanation for the increased resistance is that the battery has made the p-type semiconductor 
more negative than it used to be, making it repel any electrons from the n-type semiconductor which attempt 
to cross the depletion band. 

On the other hand, if the diode is forward biased, the depletion band is made narrower. The negative 
charge on the p-type silicon is cancelled out by the battery. The greater the voltage used, the narrower 
the depletion band becomes. Eventually, when the voltage is about 0,6 V (for silicon) the depletion band 
disappears. Once this has occurred, the diode conducts very well. 

14.3.1.1.1 The Diode 

1. What is a diode? 

2. What is a diode made of? 

3. What is the term which means that a diode is connected the 'wrong way' and little current is flowing? 

4. Why is a diode able to conduct electricity in one direction much more easily than the other? 

14.3.1.2 The Light-Emitting Diode (LED) 

A light-emitting diode (LED) is a diode device that emits light when charge flows in the correct direction 
through it. If you apply a voltage to force current to flow in the direction the LED allows, it will light up. 

Image notjinished 

Figure 14.27: Symbol for a light-emitting diode with anode and cathode labeled. 



274 CHAPTER 14. ELECTRONICS 

14.3.1.2.1 Circuit Symbols 

This notation of having two small arrows pointing away from the device is common to the schematic symbols 
of all light-emitting semiconductor devices. Conversely, if a device is light-activated (meaning that incoming 
light stimulates it), then the symbol will have two small arrows pointing toward it. It is interesting to note, 
though, that LEDs are capable of acting as light-sensing devices: they will generate a small voltage when 
exposed to light, much like a solar cell on a small scale. This property can be gainfully applied in a variety 
of light-sensing circuits. 

The colour depends on the semiconducting material used to construct the LED, and can be in the 
near-ultraviolet, visible or infrared part of the electromagnetic spectrum. 

NOTE: Nick Holonyak Jr. (1928) of the University of Illinois at Urbana-Champaign developed the 
first practical visible-spectrum LED in 1962. 



14.3.1.2.2 Light emission 

The wavelength of the light emitted, and therefore its colour, depends on the materials forming the p-n 
junction. A normal diode, typically made of silicon or germanium, emits invisible far-infrared light (so it 
can't be seen), but the materials used for an LED can emit light corresponding to near-infrared, visible or 
near-ultraviolet frequencies. 

14.3.1.2.3 LED applications 

LEDs have many uses. Some of these are given here. 

1. thin, lightweight message displays, e.g. in public information signs (at airports and railway stations, 
among other places) 

2. status indicators, e.g. on/off lights on professional instruments and consumers audio/video equipment 

3. infrared LEDs in remote controls (for TVs, VCRs, etc.) 

4. clusters of LEDs are used in traffic signals, replacing ordinary bulbs behind coloured glass 

5. car indicator lights and bicycle lighting 

6. calculator and measurement instrument displays (seven segment displays), although now mostly re- 
placed by LCDs 

7. red or yellow LEDs are used in indicator and [alpha] numeric displays in environments where night 
vision must be retained: aircraft cockpits, submarine and ship bridges, astronomy observatories, and 
in the field, e.g. night time animal watching and military field use 

8. red or yellow LEDs are also used in photographic darkrooms, for providing lighting which does not 
lead to unwanted exposure of the film 

9. illumination, e.g. flashlights (a.k.a. torches, UK), and backlighting for LCD screens 

10. signaling/emergency beacons and strobes 

11. movement sensors, e.g. in mechanical and optical computer mice and trackballs 

12. in LED printers, e.g. high-end colour printers 

LEDs offer benefits in terms of maintenance and safety. 

1. The typical working lifetime of a device, including the bulb, is ten years, which is much longer than 
the lifetimes of most other light sources. 

2. LEDs fail by dimming over time, rather than the abrupt burn-out of incandescent bulbs. 

3. LEDs give off less heat than incandescent light bulbs and are less fragile than fluorescent lamps. 

4. Since an individual device is smaller than a centimetre in length, LED-based light sources used for 
illumination and outdoor signals are built using clusters of tens of devices. 



275 

Because they are monochromatic, LED Hghts have great power advantages over white Hghts where a specific 
colour is required. UnHke the white Hghts, the LED does not need a filter that absorbs most of the emitted 
white light. Coloured fiuorescent lights are made, but they are not widely available. LED lights are inherently 
coloured, and are available in a wide range of colours. One of the most recently introduced colours is the 
emerald green (bluish green, wavelength of about 500 nm) that meets the legal requirements for traffic signals 
and navigation lights. 

NOTE: The largest LED display in the world is 36 m high, at Times Square, New York, U.S.A. 

There are applications that specifically require light that does not contain any blue component. Examples 
are photographic darkroom safe lights, illumination in laboratories where certain photo-sensitive chemicals 
are used, and situations where dark adaptation (night vision) must be preserved, such as cockpit and bridge 
illumination, observatories, etc. Yellow LED lights are a good choice to meet these special requirements 
because the human eye is more sensitive to yellow light. 

14.3.1.2.3.1 The Light Emitting Diode 

1. What is an LED? 

2. List 5 applications of LEDs. 



14.3.1.3 Transistor 

The diode is the simplest semiconductor device, made up of a p-type semiconductor and an n-type semicon- 
ductor in contact. It can conduct in only one direction, but it cannot control the size of an electric current. 
Transistors are more complicated electronic components which can control the size of the electric current 
fiowing through them. 

This enables them to be used in amplifiers. A small signal from a microphone or a radio antenna can 
be used to control the transistor. In response, the transistor will then increase and decrease a much larger 
current which fiows through the speakers. 

NOTE: One of the earliest popular uses of transistors was in cheap and portable radios. Before 
that, radios were much more expensive and contained glass valves which were fragile and needed 
replacing. In some parts of the world you can still hear people talking about their 'transistor' — 
they mean their portable radio. 

You can also use a small current to turn the transistor on and off. The transistor then controls a more 
complicated or powerful current through other components. When a transistor is used in this way it is said 
to be in switching mode as it is acting as a remotely controlled switch. As we shall see in the final sections 
of this chapter, switching circuits can be used in a computer to process and store digital information. A 
computer would not work without the millions (or billions) of transistors in it. 

There are two main types of transistor - bipolar transistors (NPN or PNP), and field effect transistors 
(FETs). Both use doped semiconductors, but in different ways. You are mainly required to know about field 
effect transistors (FETs), however we have to give a brief description of bipolar transistors so that you see 
the difference. 

14.3.1.3.1 Bipolar Transistors 

Bipolar transistors are made of a doped semiconductor 'sandwich'. In an NPN transistor, a very thin 
layer of p-type semiconductor is in between two thicker layers of n-type semiconductor. This is shown in 
Figure 14.28. Similarly an PNP transistor consists of a very thin n-type layer in between two thicker layers 
of p-type semiconductor. 



276 CHAPTER 14. ELECTRONICS 



Image notjtnished 

Figure 14.28: An NPN transistor. This is a type of bipolar transistor. 



In an NPN transistor a small current of electrons flows from the emitter (E) to the base (B). Simulta- 
neously, a much larger current of electrons flows from the emitter (E) to the collector (C). If you lower the 
number of electrons able to leave the transistor at the base (B), the transistor automatically reduces the 
number of electrons flowing from emitter (E) to collector (C). Similarly, if you increase the current of elec- 
trons flowing out of the base (B), the transistor automatically also increases the current of electrons flowing 
from emitter (E) to collector (C). The transistor is designed so that the current of electrons from emitter 
to collector {Iec) is proportional to the current of electrons from emitter to base {Ieb)- The constant of 
proportionality is known as the current gain/3. So Iec = PIeb- 

How does it do it? The answer comes from our work with diodes. Electrons arriving at the emitter 
(n-type semiconductor) will naturally flow through into the central p-type since the base-emitter junction 
is forward biased. However if none of these electrons are removed from the base, the electrons flowing into 
the base from the emitter will flll all of the available 'holes'. Accordingly, a large depletion band will be set 
up. This will act as an insulator preventing current flow into the collector as well. On the other hand, if the 
base is connected to a positive voltage, a small number of electrons will be removed by the base connection. 
This will prevent the 'holes' in the base becoming fllled up, and no depletion band will form. While some 
electrons from the emitter leave via the base connection, the bulk of them flow straight on to the collector. 
You may wonder how the electrons get from the base into the collector (it seems to be reverse biased). The 
answer is complicated, but the important fact is that the p-type layer is extremely thin. As long as there is 
no depletion layer, the bulk of the electrons will have no difficulty passing straight from the n-type emitter 
into the n-type collector. A more satisfactory answer can be given to a university student once band theory 
has been explained. 

Summing up, in an NPN transistor, a small flow of electrons from emitter (E) to base (B) allows a much 
larger flow of electrons from emitter (E) to collector (C). Given that conventional current (flowing from -|- 
to — ) is in the opposite direction to electron flow, we say that a small conventional current from base to 
emitter allows a large current to flow from collector to emitter. 

A PNP transistor works the other way. A small conventional current from emitter to base allows a much 
larger conventional current to flow from emitter to collector. The operation is more complicated to explain 
since the principal charge carrier in a PNP transistor is not the electron but the 'hole'. 

The operation of NPN and PNP transistors (in terms of conventional currents) is summarized in Fig- 
ure 14.29. 



Image notjinished 



Figure 14.29: An overview of bipolar transistors as current amplifiers. (Left) An NPN transistor. 
(Right) A PNP transistor. 



NOTE: The transistor is considered by many to be one of the greatest discoveries or inventions 
in modern history, ranking with banking and the printing press. Key to the importance of the 



277 

transistor in modern society is its ability to be produced in huge numbers using simple techniques, 
resulting in vanishingly small prices. Computer "chips" consist of millions of transistors and sell 
for Rands, with per-transistor costs in the thousandths-of-cents. The low cost has meant that 
the transistor has become an almost universal tool for non-mechanical tasks. Whereas a common 
device, say a refrigerator, would have used a mechanical device for control, today it is often less 
expensive to simply use a few million transistors and the appropriate computer program to carry out 
the same task through "brute force". Today transistors have replaced almost all electromechanical 
devices, most simple feedback systems, and appear in huge numbers in everything from computers 
to cars. 

NOTE: The transistor was invented at Bell Laboratories in December 1947 (first demonstrated 
on December 23) by John Bardeen, Walter Houser Brattain, and William Bradford Shockley, who 
were awarded the Nobel Prize in physics in 1956. 

14.3.1.3.2 The Field Effect Transistor (FET) 

To control a bipolar transistor, you control the current flowing into or out of its base. The other type of 
transistor is the field effect transistor (FET). FETs work using control voltages instead. Accordingly they 
can be controlled with much smaller currents and are much more economic to use. 

NOTE: No-one would build a computer with billions of bipolar transistors — the current in each 
transistor's base might be small, but when you add up all of the base currents in the millions of 
transistors, the computer as a whole would be consuming a great deal of electricity and making a 
great deal of heat. Not only is this wasteful, it would prevent manufacturers making a computer of 
convenient size. If the transistors were too close together, they would overheat. 



Image notjinished 

Figure 14.30 

Image notjinished 

Figure 14.31 



A field effect transistor (FET). The diagram on the top shows the semiconductor structure. The diagram 
underneath shows its circuit symbol. 

The three terminals of the FET are called the source (S), drain (D) and gate (G), as shown in . When the 
gate is not connected, a current of electrons can flow from source (S) to drain (D) easily along the channel. 
The source is, accordingly, the negative terminal of the transistor. The drain, where the electrons come out, 
is the positive terminal of the transistor. A few electrons will flow from the n-type channel into the p-type 
semiconductor of the gate when the device is manufactured. However, as these electrons are not removed 
(the gate is not connected), a depletion band is set up which prevents further flow into the gate. 



278 CHAPTER 14. ELECTRONICS 

In operation, the gate is connected to negative voltages relative to the source. This makes the p-n junction 
between gate and channel reverse-biased. Accordingly no current flows from the source into the gate. When 
the voltage of the gate is lowered (made more negative) , the depletion band becomes wider. This enlarged 
depletion band takes up some of the space of the channel. So the lower the voltage of the gate (the more 
negative it is relative to the source), the larger the depletion band. The larger the depletion band, the 
narrower the channel. The narrower the channel, the harder it is for electrons to flow from source to drain. 

The voltage of the gate is not the only factor affecting the current of electrons between the source and 
the drain. If the external circuit has a low resistance, electrons are able to leave the drain easily. If the 
external circuit has a high resistance, electrons leave the drain slowly. This creates a kind of 'traffic jam' 
which slows the passage of further electrons. In this way, the voltage of the drain regulates itself, and is 
more or less independent of the current demanded from the drain. 

Once these two factors have been taken into account, it is fair to say that the positive output voltage 
(the voltage of the drain relative to the source) is proportional to the negative input voltage (the voltage of 
the gate relative to the source). 

For this reason, the field effect transistor is known as a voltage amplifier. This contrasts with the bipolar 
transistor which is a current amplifier. 

14.3.1.3.2.1 Field Effect Transistors 

1. What are the two types of bipolar transistor? How does their construction differ? 

2. What are the three connections to a bipolar transistor called? 

3. Why are very few electrons able to fiow from emitter to collector in an NPN transistor if the base is 
not connected? 

4. Why do you think a bipolar transistor would not work if the base layer were too thick? 

5. "The bipolar transistor is a current amplifier." What does this statement mean? 

6. Describe the structure of a FET. 

7. Define what is meant by the source, drain and gate. During normal operation, what will the voltages 
of drain and gate be with respect to the source? 

8. Describe how a depletion layer forms when the gate voltage is made more negative. What controls the 
width of the depletion layer? 

9. "The field effect transistor is a voltage amplifier." What does this statement mean? 

10. The amplifier in a cheap radio will probably contain bipolar transistors. A computer contains many 
field effect transistors. Bipolar transistors are more rugged and less sensitive to interference than field 
effect transistors, which makes them more suitable for a simple radio. Why are FETs preferred for the 
computer? 



14.3.1.4 The Operational Amplifier 

The operational amplifier is a special kind of voltage amplifier which is made from a handful of bipolar or field 
effect transistors. Operational amplifiers are usually called op-amps for short. They are used extensively 
in all kinds of audio equipment (amplifiers, mixers and so on) and in instrumentation. They also have many 
other uses in other circuits - for example comparing voltages from sensors. 

Operational amplifiers are supplied on Integrated Circuits (I.C.s). The most famous operational amplifier 
I.e. is numbered 741 and contains a single operational amplifier on an integrated circuit ('chip') with eight 
terminals. Other varieties can be bought, and you can get a single integrated circuit with two or four 
'741'-type operational amplifiers on it. 

The symbol for an op-amp is shown in Figure 14.32. The operational amplifier has two input terminals 
and one output terminal. The voltage of the output terminal is proportional to the difference in voltage 
between the two input terminals. The output terminal is on the right (at the sharp point of the triangle). 
The two input terminals are drawn on the left. One input terminal (labelled with a -I- on diagrams) is called 
the non-inverting input. The other input terminal (labelled — ) is called the inverting input. The labels 



279 

+ and — have nothing to do with the way in which the operational ampHfier is connected to the power 
supply. Operational amplifiers must be connected to the power supply, but this is taken for granted when 
circuit diagrams are drawn, and these connections are not shown on circuit diagrams. Usually, when drawing 
electronic circuits, 'OV is taken to mean the negative terminal of the power supply. This is not the case 
with op-amps. For an op-amp, 'OV refers to the voltage midway between the -I- and — of the supply. 



Image notjinished 



Figure 14.32: Circuit symbol for an operational amplifier. The amplifier must also be connected to the 
-|- and — terminals of the power supply. These connections are taken for granted and not shown. 



The output voltage of the amplifier Vout is given by the formula Vout = A {V+ — V-) where A is a constant 
called the open loop gain, and V+ and V- are the voltages of the two input terminals. That said, the 
output voltage can not be less than the voltage of the negative terminal of the battery supplying it or higher 
than the positive terminal of the battery supplying it. You will notice that Vout is positive if V+ > V- and 
negative if Vj^ < V^. This is why the — input is called the inverting input: raising its voltage causes the 
output voltage to drop. 

The input resistance of an operational amplifier is very high. This means that very little current fiows 
into the input terminals during operation. 

If all of the transistors in the operational amplifier were identical then the output voltage would be zero 
if the two inputs were at equal voltages. In practice this is not quite the case, and for sensitive work a 
trimming potentiometer is connected. This is adjusted until the op-amp is zeroed correctly. 

Simple operational amplifiers require the trimming potentiometer to be built into the circuit containing 
them, and an example is shown in Figure 14.33. Other operational amplifier designs incorporate separate 
terminals for the trimming potentiometer. These special terminals are labelled offset on the manufacturer's 
diagram. The exact method of connecting the potentiometer to the offset terminals can depend on the 
design of the operational amplifier, and you need to refer to the manufacturer's data sheet for details of 
which potentiometer to use and how to connect it. 

For most commercially produced operational amplifiers (known as op-amps for short), the open loop gain 
A is very large and does not stay constant. Values of 100 000 are typical. Usually a designer would want an 
amplifier with a stable gain of smaller value, and builds the operational amplifier into a circuit like the one 
in Figure 14.33. 



Image notjinished 



Figure 14.33: An inverting amplifier built using an operational amplifier. The connections from battery 
to operational amplifier are not shown. The output voltage Vout ~ —R2Vin/Ri, as explained in the text. 
The potentiometer Rs is a trimming potentiometer. To set it, the input is connected to zero volts. The 
trimming potentiometer is then adjusted until Vout ~ 0. In all operational amplifier circuits, zero volts 
is midway between the -I- and — of the supply. 



280 CHAPTER 14. ELECTRONICS 

14.3.1.4.1 Calculating the gain of the amplifier in Figure 14.33. 

1. The input resistance of the operational ampHfier is very high. This means that very Uttle current flows 
into the inverting input of the op-amp. Accordingly, the current through resistor Ri must be almost 
the same as the current through resistor R2. This means that the ratio of the voltage across Ri to the 
voltage across R2 is the same as the ratio of the two resistances. 

2. The open loop gain A of the op-amp is very high. Assuming that the output voltage is less than a few 
volts, this means that the two input terminals must be at very similar voltages. We shall assume that 
they are at the same voltage. 

3. We want the output voltage to be zero if the input voltage is zero. Assuming that the transistors within 
the op-amp are very similar, the output voltage will only be zero for zero input voltage if V+ is very 
close to zero. We shall assume that V+ = when the trimming potentiometer is correctly adjusted. 

4. It follows from the last two statements that V- « 0, and we shall assume that it is zero. 

5. With these assumptions, the voltage across R2 is the same as Vout, and the voltage across _Ri is the 
same as Vm- Since both resistors carry the same current (as noted in point 1), we may say that the 
magnitude of Vout/Vm = R2/Ri. However, if Vin is negative, then Vout will be positive. Therefore it 
is customary to write the gain of this circuit as Vout/Vin = —R2/Ri- 



14.3.1.4.2 Operational Amplifiers 

1. What are operational amplifiers used for? 

2. Draw a simple diagram of an operational amplifier and label its terminals. 

3. Why is a trimming potentiometer needed when using an op-amp? 



281 



Solutions to Exercises in Chapter 14 

Solution to Exercise 14.1 (p. 257) 

Step 1. Xl = 27r/L = 27r x 50 x 0,003 = 0, 942f) 
Step 2. Z = VX2 + m = VO, 942^ + 52 = 5, mO, 
Step 3. I^ax = V^ax/Z = 6/5, 09 = 1, 18 A. 

Solution to Exercise 14.2 (p. 257) 

Step 1. Xc = iTzjn = 277vin3vsvin-6 = 53,0517 



2-KfC 

Step 2. Z = VXM 



/?2 



27rxl03x3xl0-'= 

\/53, 05^ + 



30^ 



60,9f2 



Solution to Exercise 14.3 (p. 265) 



Step 1. The first sensor tells us whether the room is occupied. The second sensor tells us whether the room 
is hot. The heating must come on if the room is occupied AND cold. This means that the heating 
should come on if the room is occupied AND (NOT hot). 

Step 2. To build the circuit, we first attach a NOT gate to the output of the temperature sensor. This output 
of the NOT gate will be 1 only if the room is cold. We then attach this output to an AND gate, 
together with the output from the other sensor. The output of the AND gate will only be 1 if the room 
is occupied AND the output of the NOT gate is also 1. So the heating will only come on if the room 
is in use and is cold. The circuit is shown below. 



Image notjinished 



Figure 14.34 



Solution to Exercise 14.4 (p. 265) 



Step 1. 



Image notjinished 



Figure 14.35 



Step 2. There is a column for each of the inputs, for the intermediate point C and also for the output. The 
truth table has four rows, since there are four possible inputs — 00, 01, 10 and 11. 



A 


B 


c 


Output 















1 






1 









1 


1 







Table 14.7 



282 



CHAPTER 14. ELECTRONICS 



Next we fill in the C column given that we know what a NOR gate does. 



A 


B 


c 


Output 








1 







1 







1 










1 


1 








Table 14.8 

Next, we can fill in the output, since it will always be the opposite of C (because of the NOT gate). 



A 


B 


c 


Output 








1 








1 





1 


1 








1 


1 


1 





1 



Table 14.9 

Step 3. Finally we see that this combination of gates does the same job as an OR gate. 

Solution to Exercise 14.5 (p. 266) 

Step 1. We start on the right. The '1' on the right does indeed represent one. The next '1' is in the third place 
from the right, and represents 2^ = 4. The next '1' is in the fifth place from the right and represents 
2^ = 16. Accordingly, the binary number 10101 represents 16+4+1 = 21 in denary notation. 

Solution to Exercise 14.6 (p. 266) 

Step 1. Firstly we write 12 as a sum of powers of 2, so 12 = 8+4. In binary, eight is 1000, and four is 100. 
This means that twelve = eight + four must be 1000+100 = 1100 in binary. You could also write 12 
as 1x8+1x4 + 0x2 + 0x1 = 1100 in binary. 



Chapter 15 

Electromagnetic radiation 

15.1 Wave nature and particle nature' 

15.1.1 Introduction 

This chapter will focus on the electromagnetic (EM) radiation. Electromagnetic radiation is a self- 
propagating wave in space with electric and magnetic components. These components oscillate at right 
angles to each other and to the direction of propagation, and are in phase with each other. Electromagnetic 
radiation is classified into types according to the frequency of the wave: these types include, in order of 
increasing frequency, radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation. X-rays 
and gamma rays. 

15.1.2 Particle/ wave nature of electromagnetic radiation 

K you watch a colony of ants walking up the wall, they look like a thin continuous black line. But as you 
look closer, you see that the line is made up of thousands of separated black ants. 

Light and all other types of electromagnetic radiation seems like a continuous wave at first, but when one 
performs experiments with light, one can notice that light can have both wave and particle like properties. 
Just like the individual ants, the light can also be made up of individual bundles of energy, or quanta of 
Hght. 

Light has both wave- like and particle-like properties (wave-particle duality), but only shows one or the 
other, depending on the kind of experiment we perform. A wave-type experiment shows the wave nature, 
and a particle-type experiment shows particle nature. One cannot test the wave and the particle nature at 
the same time. A particle of light is called a photon. 

Definition 15.1: Photon 

A photon is a quantum (energy packet) of light. 

The particle nature of light can be demonstrated by the interaction of photons with matter. One way in 
which light interacts with matter is via the photoelectric effect, which will be studied in detail in here^. 

15.1.2.1 Particle/wave nature of electromagnetic radiation 

1. Give examples of the behaviour of EM radiation which can best be explained using a wave model. 

2. Give examples of the behaviour of EM radiation which can best be explained using a particle model. 



^This content is available online at <http://siyavula.cnx.Org/content/m39511/l.l/>. 

^"Optical Phenomena and Properties of Matter" <http://siyavula.cnx.org/content/m33651/latest/> 



283 



284 CHAPTER 15. ELECTROMAGNETIC RADIATION 

15.1.3 The wave nature of electromagnetic radiation 

Accelerating charges emit electromagnetic waves. We have seen that a changing electric field generates a 
magnetic field and a changing magnetic field generates an electric field. This is the principle behind the 
propagation of electromagnetic waves, because electromagnetic waves, unlike sound waves, do not need a 
medium to travel through. EM waves propagate when an electric field oscillating in one plane produces a 
magnetic field oscillating in a plane at right angles to it, which produces an oscillating electric field, and so 
on. The propagation of electromagnetic waves can be described as mutual induction. 

These mutually regenerating fields travel through empty space at a constant speed of 3 x 10^ m • s^^, 
represented by c. 

Image notjinished 

Figure 15.1 



15.1.4 Electromagnetic spectrum 



Image notjinished 



Figure 15.2: The electromagnetic spectrum as a function of frequency. The different types according 
to wavelength are shown as well as everyday comparisons. 



Observe the things around you, your friend sitting next to you, a large tree across the field. How is it that 
you are able to see these things? What is it that is leaving your friend's arm and entering your eye so that 
you can see his arm? It is light. The light originally comes from the sun, or possibly a light bulb or burning 
fire. In physics, light is given the more technical term electromagnetic radiation, which includes all forms of 
light, not just the form which you can see with your eyes. 

Electromagnetic radiation allows us to observe the world around us. It is this radiation which refiects off 
of the objects around you and into your eye. The radiation your eye is sensitive to is only a small fraction 
of the total radiation emitted in the physical universe. All of the different fractions taped together make up 
the electromagnetic spectrum. 

15.1.4.1 Dispersion 

When white light is split into its component colours by a prism, you are looking at a portion of the electro- 
magnetic spectrum. 

The wavelength of a particular electromagnetic radiation will depend on how it was created. 

15.1.4.2 Wave Nature of EM Radiation 

1. List one source of electromagnetic waves. Hint: consider the spectrum diagram and look at the names 
we give to different wavelengths. 



285 

2. Explain how an EM wave propagates, with the aid of a diagram. 

3. What is the speed of light? What symbol is used to refer to the speed of light? Does the speed of light 
change? 

4. Do EM waves need a medium to travel through? 

The radiation can take on any wavelength, which means that the spectrum is continuous. Physicists broke 
down this continuous band into sections. Each section is defined by how the radiation is created, not the 
wavelength of the radiation. But each category is continuous within the min and max wavelength of that 
category, meaning there are no wavelengths excluded within some range. 

The spectrum is in order of wavelength, with the shortest wavelength at one end and the longest wave- 
length at the other. The spectrum is then broken down into categories as detailed in Table 15.1. 



Category 


Range of Wavelengths (nm) 


Range of Frequencies (Hz) 


gamma rays 


<1 


> 3 X 10^'' 


X-rays 


1-10 


3 X 10^^-3 X 10^9 


ultraviolet light 


10-400 


7,5 X 10i'*-3 X 10^^ 


visible light 


400-700 


4,3 X 10^''-7,5 X 10" 


infrared 


700-10^ 


3 X 10^^,3 X 10^^ 


microwave 


10^ - 10* 


3 X 10^-3 X 10^2 


radio waves 


>10* 


< 3 X 10^ 



Table 15.1: Electromagnetic spectrum 

Since an electromagnetic wave is still a wave, the following equation that you learnt in Grade 10 still 
applies: 



c = /-A 



(15.1) 



Exercise 15.1: EM spectrum I (Solution on p. 290.) 

Calculate the frequency of red light with a wavelength of 4, 2 x 10^^ m 

Exercise 15.2: EM spectrum II (Solution on p. 290.) 

Ultraviolet radiation has a wavelength of 200 nm. What is the frequency of the radiation? 

Examples of some uses of electromagnetic waves are shown in Table 15.2. 



Category 


Uses 


gamma rays 


used to kill the bacteria in marshmallows and to 
sterilise medical equipment 


X-rays 


used to image bone structures 


continued on next page 



286 CHAPTER 15. ELECTROMAGNETIC RADIATION 



ultraviolet light 


bees can see into the ultraviolet because flowers 
stand out more clearly at this frequency 


visible light 


used by humans to observe the world 


infrared 


night vision, heat sensors, laser metal cutting 


microwave 


microwave ovens, radar 


radio waves 


radio, television broadcasts 



Table 15.2: Uses of EM waves 

In theory the spectrum is inflnite, although realistically we can only observe wavelengths from a few 
hundred kilometers to those of gamma rays due to experimental limitations. 

Humans experience electromagnetic waves differently depending on their wavelength. Our eyes are sen- 
sitive to visible light while our skin is sensitive to infrared, and many wavelengths we do not detect at 
all. 

15.1.4.3 EM Radiation 

1. Arrange the following types of EM radiation in order of increasing frequency: infrared. X-rays, ultra- 
violet, visible, gamma. 

2. Calculate the frequency of an EM wave with a wavelength of 400 nm. 

3. Give an example of the use of each type of EM radiation, i.e. gamma rays. X-rays, ultraviolet light, 
visible light, infrared, microwave and radio and TV waves. 



15.1.5 The particle nature of electromagnetic radiation 

When we talk of electromagnetic radiation as a particle, we refer to photons, which are packets of energy. 
The energy of the photon is related to the wavelength of electromagnetic radiation according to: 

Definition 15.2: Planck's constant 

Planck's constant is a physical constant named after Max Planck. 
h= 6,626 X lO"^"* J • s 

The energy of a photon can be calculated using the formula: E = hf ot E = hj. Where E is the energy 
of the photon in joules (J), h is planck's constant, c is the speed of light, f is the frequency in hertz (Hz) and 
A is the wavelength in metres (m). 

Exercise 15.3: Calculating the energy of a photon I (Solution on p. 290.) 

Calculate the energy of a photon with a frequency of 3 x 10^* Hz 

Exercise 15.4: Calculating the energy of a photon II (Solution on p. 290.) 

What is the energy of an ultraviolet photon with a wavelength of 200 nm? 



15.1.5.1 Exercise - particle nature of EM waves 

1. How is the energy of a photon related to its frequency and wavelength? 

2. Calculate the energy of a photon of EM radiation with a frequency of 10^^ Hz. 

3. Determine the energy of a photon of EM radiation with a wavelength of 600 nm. 



287 

15.2 Penetrating ability^ 

15.2,1 Penetrating ability of electromagnetic radiation 

Different kinds of electromagnetic radiation have different penetrabilities. For example, if we take the human 
body as the object. Infrared light is emitted by the human body. Visible light is reflected off the surface of 
the human body, ultra-violet light (from sunlight) damages the skin, but X-rays are able to penetrate the 
skin and bone and allow for pictures of the inside of the human body to be taken. 

If we compare the energy of visible light to the energy of X-rays, we find that X-rays have a much higher 
energy. Usually, kinds of electromagnetic radiation with higher energy have higher penetrabilities than those 
with low energies. 

Certain kinds of electromagnetic radiation such as ultra-violet radiation. X-rays and gamma rays are 
very dangerous. Radiation such as these are called ionising radiation. Ionising radiation transfers energy as 
it passes through matter, breaking molecular bonds and creating ions. 

Excessive exposure to radiation, including sunlight. X-rays and all nuclear radiation, can cause destruction 
of biological tissue. 

15.2.1.1 Ultraviolet (UV) radiation and the skin 

UVA and UVB are different ranges of frequencies for ultraviolet (UV) light. UVA and UVB can damage 
collagen fibres which results in the speeding up skin aging. In general, UVA is the least harmful, but it can 
contribute to the aging of skin, DNA damage and possibly skin cancer. It penetrates deeply and does not 
cause sunburn. Because it does not cause reddening of the skin (erythema) it cannot be measured in the 
SPF testing. There is no good clinical measurement of the blocking of UVA radiation, but it is important 
that sunscreen block both UVA and UVB. 

UVB light can cause skin cancer. The radiation excites DNA molecules in skin cells, resulting in possible 
mutations, which can cause cancer. This cancer connection is one reason for concern about ozone depletion 
and the ozone hole. 

As a defense against UV radiation, the body tans when exposed to moderate (depending on skin type) 
levels of radiation by releasing the brown pigment melanin. This helps to block UV penetration and prevent 
damage to the vulnerable skin tissues deeper down. Suntan lotion, often referred to as sunblock or sunscreen, 
partly blocks UV and is widely available. Most of these products contain an SPF rating that describes the 
amount of protection given. This protection, however, applies only to UVB rays responsible for sunburn and 
not to UVA rays that penetrate more deeply into the skin and may also be responsible for causing cancer 
and wrinkles. Some sunscreen lotion now includes compounds such as titanium dioxide which helps protect 
against UVA rays. Other UVA blocking compounds found in sunscreen include zinc oxide and avobenzone. 

15.2.1.1.1 What makes a good sunscreen? 

• UVB protection: Padimate O, Homosalate, Octisalate (octyl salicylate), Octinoxate (octyl methoxycin- 
namate) 

• UVA protection: Avobenzone 

• UVA/UVB protection: Octocrylene, titanium dioxide, zinc oxide, Mexoryl (ecamsule) 

Another means to block UV is by wearing sun protective clothing. This is clothing that has a UPF rating 
that describes the protection given against both UVA and UVB. 

15.2.1.2 Ultraviolet radiation and the eyes 

High intensities of UVB light are hazardous to the eyes, and exposure can cause welder's flash (photo keratitis 
or arc eye) and may lead to cataracts, pterygium and Pinguecula formation. 



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288 CHAPTER 15. ELECTROMAGNETIC RADIATION 

Protective eyewear is beneficial to those who are working with or those who might be exposed to ultravi- 
olet radiation, particularly short wave UV. Given that light may reach the eye from the sides, full coverage 
eye protection is usually warranted if there is an increased risk of exposure, as in high altitude mountaineer- 
ing. Mountaineers are exposed to higher than ordinary levels of UV radiation, both because there is less 
atmospheric filtering and because of reflection from snow and ice. 

Ordinary, untreated eyeglasses give some protection. Most plastic lenses give more protection than glass 
lenses. Some plastic lens materials, such as polycarbonate, block most UV. There are protective treatments 
available for eyeglass lenses that need it which will give better protection. But even a treatment that 
completely blocks UV will not protect the eye from light that arrives around the lens. To convince yourself 
of the potential dangers of stray UV light, cover your lenses with something opaque, like aluminum foil, 
stand next to a bright light, and consider how much light you see, despite the complete blockage of the 
lenses. Most contact lenses help to protect the retina by absorbing UV radiation. 

15.2.1.3 X-rays 

While x-rays are used significantly in medicine, prolonged exposure to X-rays can lead to cell damage and 
cancer. 

For example, a mammogram is an x-ray of the human breast to detect breast cancer, but if a woman 
starts having regular mammograms when she is too young, her chances of getting breast cancer increases. 

15.2.1.4 Gamma-rays 

Due to the high energy of gamma-rays, they are able to cause serious damage when absorbed by living cells. 

Gamma-rays are not stopped by the skin and can induce DNA alteration by interfering with the genetic 
material of the cell. DNA double-strand breaks are generally accepted to be the most biologically significant 
lesion by which ionising radiation causes cancer and hereditary disease. 

A study done on Russian nuclear workers exposed to external whole-body gamma-radiation at high 
cumulative doses shows a link between radiation exposure and death from leukaemia, lung, liver, skeletal 
and other solid cancers. 

15.2.1.4.1 Cellphones and electromagnetic radiation 

Cellphone radiation and health concerns have been raised, especially following the enormous increase in the 
use of wireless mobile telephony throughout the world. This is because mobile phones use electromagnetic 
waves in the microwave range. These concerns have induced a large body of research. Concerns about 
effects on health have also been raised regarding other digital wireless systems, such as data communication 
networks. In 2009 the World Health Organisation announced that they have found a link between brain 
cancer and cellphones. 

Cellphone users are recommended to minimise radiation, by for example: 

1. Use hands-free to decrease the radiation to the head. 

2. Keep the mobile phone away from the body. 

3. Do not telephone in a car without an external antenna. 



15.2.1.5 Exercise - Penetrating ability of EM radiation 

1. Indicate the penetrating ability of the different kinds of EM radiation and relate it to energy of the 
radiation. 

2. Describe the dangers of gamma rays. X-rays and the damaging effect of ultra-violet radiation on skin 



289 



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Figure 15.3 



15.2.2 Summary 

1. Electromagnetic radiation has both a wave and a particle nature. 

2. Electromagnetic waves travel at a speed of 3 x 10® m • s~ 

3. The Electromagnetic spectrum consists of the follwing types of radiation: radio waves, microwaves, 
infrared, visible, ultraviolet. X-rays, gamma-rays. 

4. Gamma-rays have the most energy and are the most penetrating, while radio waves have the lowest 
energy and are the least penetrating. 



15.2.3 End of chapter exercise 

1. What is the energy of a photon of EM radiation with a frequency of 3 x 10® Hz? 

2. What is the energy of a photon of light with a wavelength of 660 nm? 

3. List the main types of electromagnetic radiation in order of increasing wavelength. 

4. List the main uses of: 

a. radio waves 

b. infrared 

c. gamma rays 

d. X-rays 

5. Explain why we need to protect ourselves from ultraviolet radiation from the Sun. 

6. List some advantages and disadvantages of using X-rays. 

7. What precautions should we take when using cell phones? 

8. Write a short essay on a type of electromagnetic waves. You should look at uses, advantages and 
disadvantages of your chosen radiation. 

9. Explain why some types of electromagnetic radiation are more penetrating than others. 



290 CHAPTER 15. ELECTROMAGNETIC RADIATION 

Solutions to Exercises in Chapter 15 

Solution to Exercise 15.1 (p. 285) 

Step 1. We use the formula: c = /A to calculate frequency. The speed of light is a constant 3 x 10*m/s. 



(15.2) 



Solution to Exercise 15.2 (p. 285) 

Step 1. Recall that all radiation travels at the speed of light (c) in vacuum. Since the question does not specify 
through what type of material the Ultraviolet radiation is traveling, one can assume that it is traveling 
through a vacuum. We can identify two properties of the radiation - wavelength (200 nm) and speed 
(c). 

Step 2. 

c = /A 

(15.3) 



c = 


/A 


3 X 10® = 


/ x4,2 X 10"^ 


/ = 


7,14 X 10"Hz 





c = /A 




3 X 10® = / X 200 X 10"^ 




/ = 1.5 X 10^'^ Hz 


Solution to Exercise 15.3 (p. 286) 




Step 1. 






E = hf 




= 6,6 X 10-3^ X 3x 10^® 




2x10^15 J 



(15.4) 



Solution to Exercise 15.4 (p. 286) 

Step 1. We are required to calculate the energy associated with a photon of ultraviolet light with a wavelength 
of 200 nm. 
We can use: 

E = h^ (15.5) 

A 

Step 2. 

E = hf 

= (6,626x10-^^) -^^121^ (15.6) 

9,939x10-1° J 



Chapter 16 

Optical phenomena and properties of 
matter 



16.1 Transmission and scattering of light' 

16.1.1 Introduction 

For centuries physicists argued about whether light was a particle or a wave. It was assumed that light could 
only be one or the other, but not both. 

In earlier chapters on waves and optics, you studied how light or other electromagnetic radiation propa- 
gates like a wave. The wave nature of light was demonstrated by the propagation of light in examples such 
as diffraction, interference, and polarisation of light. 

You also saw in Electromagnetic Radiation how light sometimes behaves as a particle. This chapter 
looks at evidence supporting the particle model of light. The idea that light can have both wave and particle 
properties was one of the most important discoveries of the twentieth century. 

16.1.2 The transmission and scattering of light 

16.1.2.1 Energy levels of an electron 

We have seen that the electrons in an atom have different energy levels. When the electron receives enough 
energy, it can jump up to a higher energy level. This is called 'exciting' the electron. When the electron in 
a high energy level sheds some energy, it drops to a lower energy level. 

We have also seen that the energy associated with light at a specific wavelength is given by: 

i^=y. (16.1) 

In the particle model of light, this means that each packet of light (photon) at a wavelength A has energy: 

E=^. (16.2) 

For the electron to receive energy, it absorbs a photon and gets its energy. When an electron loses energy 
to drop to a lower level, it emits (gives off) a photon with that energy. 



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292 



CHAPTER 16. OPTICAL PHENOMENA AND PROPERTIES OF MATTER 



16.1.2.2 Interaction of light w^ith metals 

When Ught encounters or passes through a material, the photons of the Ught interact with the atoms or 
molecules of the material. Depending on the strength of the interactions and how often they happen, the 
light will pass through the material or be scattered in some other direction. 

Each wavelength of light relates to a particular energy, and the closer that energy is to the energy 
difference between two of the levels of the atom, the likelier the photon is to interact with the atom. 

When visible or ultraviolet (UV) radiation shines on a metal, the photons are absorbed by the electrons 
in the metal. The electrons are then excited up to a higher energy level. When an electron returns to a 
lower energy level, another photon is emitted. This is how light appears to be reflected off a metal surface. 

In previous chapters, you have studied geometrical optics, which tells us what happens to rays of light 
when they are reflected off a surface or refracted through a lens. That tells us what happens to light rays, 
made up of many photons, on a large scale. If you look at a smaller level, i.e. on a microscopic scale, then 
reflection and refraction happen by all the photons interacting with the atoms of the lens or mirror. The 
photons get absorbed and re-emitted many times before emerging as the finals rays of light that we see. 

Scattering of light is responsible for many effects in everyday life. We see that certain materials are red 
or blue, for example, since they contain materials that have energy level differences that correspond to the 
energies of the photons that make up red or blue light. These materials then reflect the red or blue light 
and absorb the other wavelengths in the visible spectrum. White objects reflect photons of all wavelengths 
in the visual spectrum, while black objects absorb these photons. 

NOTE: Because a truly black object absorbs all the visual wavelengths of light, and does not 
re-emit photons at visual wavelengths, we can say that 'black' is not a colour itself, but rather a 
lack of colour! Also, since black objects absorb visual light, they heat up more than objects of 
other colours which reflect light at certain wavelengths. 



16.1.2.2.1 Investigation : Reflection and absorption 

Aim: 

Investigate the interaction of light with differently coloured metal objects 

Apparatus: 

Find some differently coloured metal objects (at least 5) which will not be damaged if left in the sun for 
15 minutes. Make sure to include at least one white item and one black item. 

Method: 

At the start of your lesson set out your objects in direct sunlight. Leave them there for around 15 minutes. 

Alternatively, if it is a sunny day, you can use your teachers' cars for this experiment - as long as there 
are some cars of different colours and they have been standing in direct sunlight for the same length of time! 

After 15 minutes is up, touch each of the items/cars (be careful not to burn yourself!) and compare their 
temperatures (rate them from 1 to 5 with 1 being cold and 5 being very hot) in a table such as the example 
table below: 



Object 


Colour 


Temperature rating 


e.g. car 1 


e.g. red 


e.g. 3 









Table 16.1 



Questions: 



1. Which object was the hottest and what was it made of? 

2. Which object was the coolest and what was it made of? 



293 

3. How did the temperatures of the black and white objects compare to each other? (which one was 
hotter and which was cooler?) 

4. Try to explain the reasons for the different temperatures of the objects with respect to their colours 
and the materials of which they are made. 

Metals generally reflect most wavelengths of visible light, but they will reflect the light in a certain direction, 
given by the laws of reflection in geometrical optics. This is different to most materials, like wood or fabric, 
which reflect light in all directions. Metals have this property since they have electrons that are not bound to 
atoms and can move freely through the metal. This is unlike most other materials that have their electrons 
bound closely to the atoms. These free electrons in metals can then absorb and reflect photons of a wide 
range of energies. 

Ultraviolet light (which has shorter wavelengths than visible light) will pass through some substances, 
such as many plastics, because they do not have the right energy levels to absorb it and re-emit it. X-rays 
(also short wavelengths) will also pass through most materials, since the energies of X-rays correspond to 
the energy levels of atomic nuclei. Such nuclei are much smaller than atoms, so it is much less likely for an 
X-ray to hit a nucleus instead of the whole atom. 

Most materials will absorb infrared radiation (longer wavelengths than visible light), since the energies 
of that radiation often correspond to rotational or vibrational energy levels of molecules. 




incoming outgoing incoming outgoing 

photon ,-^fc^, photon photon photon 



Photon Transmission Plioton Scattering 

Figure 16.1: Diagrams of photon transmission (left) and scattering (right). 

16.1.2.3 Why is the sky blue? 

The sun emits light at many different wavelengths, including all of the visible wavelengths. Light which is 
made up of all the visible wavelengths appears white. So what causes the sky to look blue? 

The atmosphere consists of molecules of different gases as well as tiny dust grains. Light from the sun 
scatters off the molecules in the air (called Rayleigh scattering). The chance that the light will scatter off 
the gas molecules is higher for shorter wavelengths. The short wavelengthhlue light is therefore scattered 
more than the other colours. At noon, when the light from the sun is coming straight down (see the 
picture), the scattered blue light reaches your eyes from all directions and so the sky appears blue. The 
other wavelengths do not get scattered much and therefore miss your line of sight and are not seen. At 
sunrise or sunset, the direction of the light coming from the sun is now straight towards your eyes (see the 
picture). Therefore the scattered blue light can't be seen because it is scattered out of your line of sight. 
The redder colours (oranges and reds) can now be seen because they are not scattered as much and still fall 
in your line of sight. 



294 CHAPTER 16. OPTICAL PHENOMENA AND PROPERTIES OF MATTER 



Image notjinished 



Figure 16.2 



Image notjinished 

Figure 16.3 



16.1.2.3.1 Transmission and scattering of light 

1. Explain how visible light is reflected from metals. 

2. Explain why the sky is blue. 



16.2 Photoelectric effect" 
16.2.1 The photoelectric efFect 

Around the turn of the twentieth century, it was observed by a number of physicists (including Hertz, 
Thomson and Von Lenard) that when light was shone on a metal, electrons were emitted by the metal. This 
is called the photoelectric effect, {photo- for light, electric- for the electron.) 

Definition 16.1: The photoelectric effect 

The photoelectric effect is the process whereby an electron is emitted by a metal when light shines 
on it. 

At that time, light was thought to be purely a wave. Therefore, physicists thought that if a more intense 
(i.e. brighter) light was shone on a metal, then the electrons would be knocked out with greater kinetic 
energies than if a faint light was shone on them. However, Von Lenard observed that this did not happen 
at all. The intensity of the light made no difference to the kinetic energy of the emitted electrons! Also, it 
was observed that the electrons were emitted immediately when light was shone on the metal - there was no 
time delay. 

Einstein solved this problem by proposing that light is made up of packets of energy called quanta (now 
called photons) which interacted with the electrons in the metal like particles instead of waves. Each incident 
photon would transfer all its energy to one electron in the metal. For a speciflc colour of light (i.e. a certain 
wavelength or frequency), the energy of the photons is given hy E = hf = hc/X, where h is Planck's constant. 
The energy needed to knock an electron out of the metal is called the work function (symbol </>) of the 
metal. Therefore, the amount of energy left over as the kinetic energy (Ek) of the emitted electron would be 
the difference between the incoming photon's energy and the energy needed to knock out the electron (work 
function of the metal) : 

Ek = hf-^ (16.3) 



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295 

Increasing the intensity of the light (i.e. making it brighter) did not change the wavelength of the light and 
therefore the electrons would be emitted with the same kinetic energy as before! This solved the paradox 
and showed that light has both a wave nature and a particle nature. Einstein won the Nobel prize for 
this quantum theory and his explanation of the photoelectric effect. 

Increasing the intensity of the light actually means increasing the number of incident photons. Therefore, 
since each photon only gives energy to one electron, more incident photons means more electrons would be 
knocked out of the metal, but their kinetic energies would be the same as before. 




Incoming radiation 




Electrons knocked out 



Sea of electrons 
inside the metal 
waiting to be set free 




Figure 16.4: The photoelectric effect: Incoming photons on the left hit the electrons inside the metal 
surface. The electrons absorb the energy from the photons, and are ejected from the metal surface. 



NOTE: The photoelectric effect was first observed in the experiments of Heinrich Hertz in 1887. 
In 1899 J.J. Thomson proved that it was electrons that were emitted. The photoelectric effect was 
theoretically explained by Albert Einstein in 1905. 



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Figure 16.5 



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CHAPTER 16. OPTICAL PHENOMENA AND PROPERTIES OF MATTER 



The discovery and understanding of the photoelectric effect was one of the major breakthroughs in 
science in the twentieth century as it provided concrete evidence of the particle nature of light. It overturned 
previously held views that light was composed purely of a continuous transverse wave. On the one hand, 
the wave nature is a good description of phenomena such as diffraction and interference for light, and on 
the other hand, the photoelectric effect demonstrates the particle nature of light. This is now known as the 
'dual-nature' of light, (dual means two) 

While solving problems we need to decide for ourselves whether we should consider the wave property or 
the particle property of light. For example, when dealing with interference and diffraction, light should be 
treated as a wave, whereas when dealing with photoelectric effect we consider the particle nature. 

16.2.1.1 Applications of the photoelectric effect 

We have learnt that a metal contains electrons that are free to move between the valence and conduction 
bands. When a photon strikes the surface of a metal, it gives all its energy to one electron in the metal. 

• If the photon energy is equal to the energy between two energy levels then the electron is excited to 
the higher energy level. 

• If the photon energy is greater than or equal to the work function (energy needed to escape from the 
metal), then the electron is emitted from the surface of the metal (the photoelectric effect). 

The work function is different for different elements. The smaller the work function, the easier it is for 
electrons to be emitted from the metal. Metals with low work functions make good conductors. This is 
because the electrons are attached less strongly to their surroundings and can move more easily through 
these materials. This reduces the resistance of the material to the flow of current i.e. it conducts well. Table 
16.2 shows the work functions for a range of elements. 



Element 


Work Function (J) 


Aluminium 


6,9 X 10^1^ 


Beryllium 


8,0 X 10-1^ 


Calcium 


4,6 X 10"^^ 


Copper 


7,5 X 10"^^ 


Gold 


8,2 X W-^^ 


Lead 


6,9x 10-19 


Silicon 


1,8 X 10^19 


Silver 


6,9x 10-19 


Sodium 


3,7x 10-19 



Table 16.2: Work functions of selected elements determined from the photoelectric effect. (From the 

Handbook of Chemistry and Physics.) 



NOTE: The electron volt (eV) is the kinetic energy gained by an electron passing through a 
potential difference of one volt {IV). A volt is not a measure of energy, but the electron volt is 
a unit of energy. When you connect a 1.5 F battery to a circuit, you can give 1.5 eV of energy to 
every electron. 

Exercise 16.1: The photoelectric effect - I (Solution on p. 312.) 

Ultraviolet radiation with a wavelength of 250 nm is incident on a silver foil (work function cj) = 
6, 9 X 10~i9)_ What is the maximum kinetic energy of the emitted electrons? 



297 



Exercise 16.2: The photoelectric effect - II (Solution on p. 312.) 

If we were to shine the same uhraviolet radiation (/ = 1, 2 x 10^"'' Hz), on a gold foil (work function 
= 8, 2 X 10~^^ J), would any electrons be emitted from the surface of the gold foil? 



16.2.1.1.1 Units of energy 

When dealing with calculations at a small scale (like at the level of electrons) it is more convenient to use 
different units for energy rather than the joule (J). We define a unit called the electron-volt (eV) as the 
kinetic energy gained by an electron passing through a potential difference of one volt. 

E = qxV (16.4) 

where q is the charge of the electron and V is the potential difference applied. The charge of 1 electron is 
1, 6 X 10~^^ C, so 1 eV is calculated to be: 

1 eV = (l, 610"^^ C X 1 y) = 1, 6 X 10"^^ J (16.5) 

You can see that 1,6 x 10~^^ J is a very small amount of energy and so using electron-volts (eV) at this 
level is easier. 

Hence, leV = 1.6 x 10^19 J which means that 1 J = 6.241 x 10^*^ eV 

16.2.1.2 Real-life applications 

16.2.1.2.1 Solar Cells 

The photo-electric effect may seem like a very easy way to produce electricity from the sun. This is why 
people choose to make solar panels out of materials like silicon, to generate electricity. In real-life however, 
the amount of electricity generated is less than expected. This is because not every photon knocks out an 
electron. Other processes such as reflection or scattering also happen. This means that only a fraction sa 10% 
(depends on the material) of the photons produce photoelectrons. This drop in efficiency results in a lower 
current. Much work is being done in industry to improve this efficiency so that the panels can generate as 
high a current as possible, and create as much electricity as possible from the sun. But even these smaller 
electrical currents are useful in applications like solar-powered calculators. 

16.2.1.2.1.1 The photoelectric effect 

1. Describe the photoelectric effect. 

2. List two reasons why the observation of the photoelectric effect was significant. 

3. Refer to Table 16.2: If I shine ultraviolet light with a wavelength of 288 nm onto some aluminium foil, 
what would be the kinetic energy of the emitted electrons? 

4. I shine a light of an unknown wavelength onto some silver foil. The light has only enough energy to 
eject electrons from the silver foil but not enough to give them kinetic energy. (Refer to Table 16.2 
when answering the questions below:) 

a. If I shine the same light onto some copper foil, would electrons be ejected? 

b. If I shine the same light onto some silicon, would electrons be ejected? 

c. If I increase the intensity of the light shining on the silver foil, what happens? 

d. If I increase the frequency of the light shining on the silver foil, what happens? 



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CHAPTER 16. OPTICAL PHENOMENA AND PROPERTIES OF MATTER 



16.3 Emission and absorption spectra* 
16.3.1 Emission and absorption spectra 

16.3.1.1 Emission Spectra 

You have learnt previously about the structure of an atom. The electrons surrounding the atomic nucleus 
are arranged in a series of levels of increasing energy. Each element has its own distinct set of energy levels. 
This arrangement of energy levels serves as the atom's unique fingerprint. 

In the early 1900s, scientists found that a liquid or solid heated to high temperatures would give off a 
broad range of colours of light. However, a gas heated to similar temperatures would emit light only at 
certain specific colours (wavelengths). The reason for this observation was not understood at the time. 

Scientists studied this effect using a discharge tube. 



Photons of 
various energies 





Current 



Figure 16.6: Diagram of a discharge tube. The tube is filled with a gas. When a high enough voltage 
is applied across the tube, the gas ionises and acts like a conductor, allowing a current to flow through 
the circuit. The current excites the atoms of the ionised gas. When the atoms fall back to their ground 
state, they emit photons to carry off the excess energy. 



A discharge tube (Figure 16.6) is a glass gas-filled tube with a metal plate at both ends. If a large enough 
voltage difference is applied between the two metal plates, the gas atoms inside the tube will absorb enough 
energy to make some of their electrons come off i.e. the gas atoms are ionised. These electrons start moving 
through the gas and create a current, which raises some electrons in other atoms to higher energy levels. 
Then as the electrons in the atoms fall back down, they emit electromagnetic radiation (light). The amount 
of light emitted at different wavelengths, called the emission spectrum, is shown for a discharge tube filled 



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299 

with hydrogen gas in Figure 16.7 below. Only certain wavelengths (i.e. colours) of light are seen as shown 
by the thick black lines in the picture. 



410 nm 434nm 

.1 I 



400 nm 



486 nm 

J, 



656 nm 

J_ 



500 nm 



600 nm 



700 nm 



Figure 16.7: Diagram of the emission spectrum of hydrogen in the visible spectrum. Four lines are 
visible, and are labeled with their wavelengths. The three lines in the 400-500 nm range are in the blue 
part of the spectrum, while the higher line (656 nm) is in the red/orange part. 



Eventually, scientists realized that these lines come from photons of a specific energy, emitted by electrons 
making transitions between specific energy levels of the atom, shows an example of this happening. When 
an electron in an atom falls from a higher energy level to a lower energy level, it emits a photon to carry 
off the extra energy. This photon's energy is equal to the energy difference between the two energy levels. 
As we previously discussed, the frequency of a photon is related to its energy through the equation E = hf. 
Since a specific photon frequency (or wavelength) gives us a specific colour, we can see how each coloured 
line is associated with a specific transition. 



Image notjtnished 



Figure 16.8: In this diagram are shown some of the electron energy levels for the hydrogen atom. The 
arrows show the electron transitions from higher energy levels to lower energy levels. The energies of 
the emitted photons are the same as the energy difference between two energy levels. You can think of 
absorption as the opposite process. The arrows would point upwards and the electrons would jump up 
to higher levels when they absorp a photon of the right energy. 



Visible light is not the only kind of electromagnetic radiation emitted. More energetic or less energetic 
transitions can produce ultraviolet or infrared radiation. However, because each atom has its own distinct 
set of energy levels (its fingerprint!), each atom has its own distinct emission spectrum. 



16.3.1.2 Absorption spectra 

As you know, atoms do not only emit photons; they also absorb photons. If a photon hits an atom and the 
energy of the photon is the same as the gap between two electron energy levels in the atom, then the electron 
can absorb the photon and jump up to the higher energy level. If the atom has no energy level differences 
that equal the incoming photon's energy, it cannot absorb the photon, and can only scatter it. 

Using this effect, if we have a source of photons of various energies we can obtain the absorption 
spectra for different materials. To get an absorption spectrum, just shine white light on a sample of the 
material that you are interested in. White light is made up of all the different wavelengths of visible light 
put together. In the absorption spectrum, the energy levels corresponding to the absorbed photons show up 



300 



CHAPTER 16. OPTICAL PHENOMENA AND PROPERTIES OF MATTER 



as black lines because the photons of these wavelengths have been absorbed and don't show up. Because of 
this, the absorption spectrum is the exact inverse of the emission spectrum. Look at the two figures below. 
In Figure 16.9 you can see the emission lines of hyrodrogen. Figure 16.10 shows the absorption spectrum. 
It is the exact opposite of the emission spectrum! Both emission and absorption techniques can be used to 
get the same information about the energy levels of an atom. 



Image notjinished 



Figure 16.9: Emission spectrum of Hydrogen. 



Image notjinished 



Figure 16.10: Absorption spectrum of Hydrogen. 



Exercise 16.3: Absorption (Solution on p. 312.) 

I have an unknown gas in a glass container. I shine a bright white light through one side of 
the container and measure the spectrum of transmitted light. I notice that there is a black line 
{absorption line) in the middle of the visible red band at 642 nm. I have a hunch that the gas might 
be hydrogen. If I am correct, between which 2 energy levels does this transition occur? (Hint: look 
at Figure 16.8 and the transitions which are in the visible part of the spectrum.) 



16.3.1.3 Colours and energies of electromagnetic radiation 

We saw in the explanation for why the sky is blue that different wavelengths or frequencies of light correspond 
to different colours of light. The table below gives the wavelengths and colours of light in the visible spectrum: 



Colour 


Wavelength range (nm) 


violet 


390 - 455 


blue 


455 - 492 


green 


492 - 577 


yellow 


577 - 597 


orange 


597 - 622 


red 


622 - 780 



Table 16.3: Colours and wavelengths of light in the visible spectrum. 



301 

We also know that the energy of a photon of Hght can be found from: 

E = hf = ^ (16.6) 

Therefore if we know the frequency or wavelength of light, we can calculate the photon's energy and vice 
versa. 

16.3.1.3.1 Investigation : Frequency, wavelength and energy relation 

Refer to Table 16.3: Copy the table into your workbook and add two additional columns. 

1. In the first new column write down the lower and upper frequencies for each colour of light. 

2. In the second column write down the energy range (in Joules) for each colour of light. 

Questions 

1. Which colour of visible light has the highest energy photons? 

2. Which colour of visible light has the lowest energy photons? 

Discharge lamps (sometimes incorrectly called neon lights) use the spectra of various elements to produce 
lights of many colours. 



Image notjinished 

Figure 16.11 



run demo^ 

Exercise 16.4: Colours of light (Solution on p. 313.) 

A photon of wavelength 500 nm is emitted by a traffic light. 

1. What is the energy of the photon? 

2. What is the frequency of the photon? 

3. Use Table 16.3 to determine the colour of the light. 

Exercise 16.5: Colours and energies of light (Solution on p. 313.) 

I have some sources which emit light of the following wavelengths: 

1. 400 nm, 

2. 580 nm, 

3. 650 nm, 

4. 300 nm. 

What are the colours of light emitted by the sources (see Table 16.3)? Which source emits photons 
with the highest energy and which with the lowest energy? 



16.3.1.4 Applications of emission and absorption spectra 

The study of spectra from stars and galaxies in astronomy is called spectroscopy. Spectroscopy is a tool 
widely used in astronomy to learn different things about astronomical objects. 



^http://phet. Colorado. edu/sims/discharge-lamps/discharge-lampsen.jnlp 



302 CHAPTER 16. OPTICAL PHENOMENA AND PROPERTIES OF MATTER 

16.3.1.4.1 Identifying elements in astronomical objects using their spectra 

Measuring the spectrum of Hght from a star can tell astronomers what the star is made of! Since each 
element emits or absorbs light only at particular wavelengths, astronomers can identify what elements are 
in the stars from the lines in their spectra. From studying the spectra of many stars we know that there are 
many different types of stars which contain different elements and in different amounts. 

16.3.1.4.2 Determining velocities of galaxies using spectroscopy 

You have already learned in Chapter about the Doppler effect and how the frequency (and wavelength) of 
sound waves changes depending on whether the object emitting the sound is moving towards or away from 
you. The same thing happens to electromagnetic radiation (light). If the object emitting the light is moving 
towards us, then the wavelength of the light appears shorter (called blue-shifted). If the object is moving 
away from us, then the wavelength of its light appears stretched out (called red-shifted). 

The Doppler effect affects the spectra of objects in space depending on their motion relative to us on 
the earth. For example, the light from a distant galaxy, which is moving away from us at some velocity, 
will appear red-shifted. This means that the emission and absorption lines in the galaxy's spectrum will be 
shifted to a longer wavelength (lower frequency). Knowing where each line in the spectrum would normally 
be if the galaxy was not moving, and comparing to their red-shifted positions, allows astronomers to precisely 
measure the velocity of the galaxy relative to the earth! 

16.3.1.4.3 Global warming and greenhouse gases 

The sun emits radiation (light) over a range of wavelengths which are mainly in the visible part of the 
spectrum. Radiation at these wavelengths passes through the gases of the atmosphere to warm the land and 
the oceans below. The warm earth then radiates this heat at longer infrared wavelengths. Carbon-dioxide 
(one of the main greenhouse gases) in the atmosphere has energy levels which correspond to the infrared 
wavelengths which allow it to absorb the infrared radiation. It then also emits at infrared wavelengths in all 
directions. This effect stops a large amount of the infrared radiation getting out of the atmosphere, which 
causes the atmosphere and the earth to heat up. More radiation is coming in than is getting back out. 



Image notjinished 

Figure 16.12 



Therefore increasing the amount of greenhouse gases in the atmosphere increases the amount of trapped 
infrared radiation and therefore the overall temperature of the earth. The earth is a very sensitive and com- 
plicated system upon which life depends and changing the delicate balances of temperature and atmospheric 
gas content may have disastrous consequences if we are not careful. 

16.3.1.4.3.1 Investigation : The greenhouse effect 

In pairs try to find the following information (e.g. in books, on the internet) and report back to the class in 
a 5 minute presentation which includes the following: 

1. What other gases besides carbon dioxide are responsible for the greenhouse effect? 

2. Where do greenhouse gases come from? (are they human-made or natural?) 

3. Investigate one serious side-effect which could arise if the earth's temperature were to go up significantly. 
Present some ways in which this effect could be avoided. 



303 

16.3.1.4.3.2 Emission and absorption spectra 

1. Explain how atomic emission spectra arise and how they relate to each element on the periodic table. 

2. How do the lines on the atomic spectrum relate to electron transitions between energy levels? 

3. Explain the difference between atomic absorption and emission spectra. 

4. Describe how the absorption and emission spectra of the gases in the atmosphere give rise to the 
Greenhouse Effect. 

5. Using Table 16.3 calculate the frequency range for yellow light. 

6. What colour is the light emitted by hydrogen when an electron makes the transition from energy level 
5 down to energy level 2? (Use Figure 16.8 to find the energy of the released photon.) 

7. I have a glass tube filled with hydrogen gas. I shine white light onto the tube. The spectrum I then 
measure has an absorption line at a wavelength of 474 nm. Between which two energy levels did the 
transition occur? (Use Figure 16.8 in solving the problem.) 



16.4 Lasers' 
16.4,1 Lasers 

A laser is a device that produces a special type of light: all the laser photons are identical! They all have 
the same wavelength (and frequency), amplitude and phase. Since they all have the same wavelength, this 
means they all have the same colour and the light is called monochromatic. {Note:mono means "one" or 
"single" and chromatic means "colour".) This is very different to most other light sources which produce 
light with a range of wavelengths (e.g. white light from the sun consists of all the visible wavelengths.) 

Laser light is highly directional and can be focused very well. This focus allows laser beams to be used 
over long distances, and to pack a lot of energy into the beam while still requiring reasonably small amounts 
of energy to be generated. Each centimetre of a typical laser beam contains many billions of photons. These 
special properties of laser light come from the way in which the laser photons are created and the energy 
levels of the material that makes up the laser. These properties make laser light extremely useful in many 
applications from CD players to eye surgery. 

The term LASER stands for Light Amplification by the Stimulated Emission of Radiation. This 
stimulated emission is different to the spontaneous emission already discussed earlier. Let's review the 
absorption and emission processes which can occur in atoms. 



Image notjinished 

Figure 16.13 



Absorption: As you can see in the picture above, absorption happens when an electron jumps up 
to a higher energy level by absorbing a photon which has an energy equal to the energy difference 
between the two energy levels. 

Spontaneous emission: Spontaneous emission is when an electron in a higher energy level drops 
down to a lower energy level and a photon is emitted with an energy equal to the energy difference 
between the two levels. There is no interference in this process from outside factors. Usually sponta- 
neous emission happens very quickly after an electron gets into an excited state. In other words, the 
lifetime of the excited state is very short (the electron only stays in the high energy level for a very 



^This content is available online at <http://siyavula.cnx.Org/content/m39557/l.l/>. 



304 CHAPTER 16. OPTICAL PHENOMENA AND PROPERTIES OF MATTER 

short time). However, there are some excited states where an electron can remain in the higher energy 
level for a longer time than usual before dropping down to a lower level. These excited states are called 
metastable states. 
• Stimulated emission: As the picture above shows, stimulated emission happens when a photon 
with an energy equal to the energy difference between two levels interacts with an electron in the higher 
level. This stimulates the electron to emit an identical photon and drop down to the lower energy level. 
This process results in two photons at the end. 

Definition 16.2: Spontaneous Emission 

Spontaneous emission occurs when an atom is in an unstable excited state and randomly decays to 
a less energetic state, emitting a photon to carry off the excess energy. The unstable state decays 
in a characteristic time, called the lifetime. 

Definition 16.3: Meta-stable state 

A meta-stable state is an excited atomic state that has an unusually long lifetime, compared to the 
lifetimes of other excited states of that atom. While most excited states have lifetimes measured 
in microseconds and nanoseconds (10~^ s and 10~^ s), meta-stable states can have lifetimes of 
milliseconds (10~^ s) or even seconds. 

Definition 16.4: Stimulated emission 

Stimulated emission occurs when a photon interacts with an excited atom, causing the atom to 
decay and emit another identical photon. 

16.4.1.1 How a laser works 

A laser works by a process called stimulated emission - as you can tell from what 'laser' stands for! You 
can imagine that stimulated emission can lead to more and more identical photons being released in the 
following way: Imagine we have an electron in an excited metastable state and it drops down to the ground 
state by emitting a photon. If this photon then travels through the material and meets another electron in 
the metastable excited state this will cause the electron to drop down to the lower energy level and another 
photon to be emitted. Now there are two photons of the same energy. If these photons then both move 
through the material and each interacts with another electron in a metastable state, this will result in them 
each causing an additional photon to be released, i.e. from 2 photons we then get 4, and so on! This is how 
laser light is produced. 



305 



Spontaneous Emission 

"metastable" 



excited '^^®'' 



Outside A-'"' After^ ^"\ 

Energy — ► I Some ^ 

Source I Time ▼ 



ground ground 

state state 

first step second step 

Figure 16.14: Spontaneous emission is a two step process, as shown here. First, energy from an external 
source is applied to an atom in the laser medium, raising its energy to an excited (metastable) state. 
After some time, it will decay back down to its ground state and emit the excess energy in the form of 
a photon. This is the first stage in the formation of a laser beam. 



306 CHAPTER 16. OPTICAL PHENOMENA AND PROPERTIES OF MATTER 



Stimulated Emission 



laser 
photon 


excited 

state 


^ excited'-' ^ r\ r\ r 
"> state www 

2 laser 
w ^ photons 


first step 


ground 
state 


ground 
state 

second step 



Figure 16.15: Stimulated emission is also a two step process, as shown here. First, a laser photon 
encounters an atom that has been raised to an excited state, just like in the case of spontaneous emission. 
The photon then causes the atom to decay to its ground state and emit another photon identical to the 
incoming photon. This is the second step in the creation of a laser beam. It happens many, many times 
as the laser photons pass through the optical cavity until the laser beam builds up to full strength. 



This can only happen if there are many electrons in a metastable state. If most of the electrons are in 
the ground state, then they will just absorb the photons and no extra photons will be emitted. However, if 
more electrons are in the excited metastable state than in the ground state, then the process of stimulated 
emission will be able to continue. Usually in atoms, most of the electrons are in the lower energy levels and 
only a few are in excited states. When most of the electrons are in the excited metastable state and only a 
few are in the ground state, this is called population inversion (the populations in the excited and ground 
states are swapped around) and this is when stimulated emission can occur. To start off the process, the 
electrons first have to be excited up into the metastable state. This is done using an external energy source. 

Definition 16.5: Population inversion 

Population inversion is when more atoms are in an excited state than in their ground state. It is 
a necessary condition to sustain a laser beam, so that there are enough excited atoms that can be 
stimulated to emit more photons. 



Image notjinished 

Figure 16.16 



Therefore, materials used to make laser light must must have metastable states which can allow population 
inversion to occur when an external energy source is applied. Some substances which are used to make lasers 



307 

are listed in Table 16.4. You can see that gases (such as Helium-Neon mixture), liquids (such as dyes), and 
solids (such as the precious stone ruby) are all used to make lasers. 



Material 


Type 


Wavelength 


Uses 


HeHum-Neon 


gas 


632,8 nm 


scientific research, 
holography 


Argon ion 


gas 


488,0 nm 


medicine. 


Carbon dioxide 


gas 


10,6 /xm 


industry (cutting, weld- 
ing), surgery 


HeHum-Cadmium 


vapor 


325 nm 


printing, scientific re- 
search 


Ruby 


solid-state 


694,3 nm 


holography 


Neodymium YAG (Yt- 
trium Aluminium Gar- 
net) 


solid-state 


1,064 ^m 


industry, surgery, re- 
search 


Titanium-Sapphire 


solid-state 


650-1100 nm 


research 


Laser diode 


semiconductor 


375-1080 nm 


telecommunications, in- 
dustry. 








printing, CD players, 
laser pointers 



Table 16.4: A selection of different lasers. The laser material and general type of each laser is given, along 

with typical wavelengths of the laser light they create. Examples of real-world applications are also given. 

All these materials allow a population inversion to be set up. 

NOTE: The first working laser, using synthetic ruby as the laser material, was made by Theodore 
H. Maiman at Hughes Research Laboratories in Malibu, California. Later in the same year the 
Iranian physicist Ali Javan, together with William Bennet and Donald Herriot, made the first gas 
laser using helium and neon. Javan received the Albert Einstein Award in 1993. 



16.4.1.2 A simple laser 

A laser consists of a number of different parts that work together to create the laser beam. Figure 16.17 
shows the different parts of the laser, while Figure 16.18 shows how they create the laser beam. 



308 



CHAPTER 16. OPTICAL PHENOMENA AND PROPERTIES OF MATTER 



External 

Energy 

Source 



1 



Laser Medium 



Laser Beam 



Full Mirror 



Partially-Silvered 
Mirror 



Optical Cavity 



Figure 16.17: Diagram of a laser showing the main components. 



External 

Energy 

Source 



f 






(D 







Figure 16.18: Diagram of a laser showing the process of creating a laser beam. (1) A source of external 
energy is applied to the laser medium, raising the atoms to an excited state. (2) An excited atom decays 
though spontaneous emission, emitting a photon. (3) The photon encounters another excited atom and 
causes it to decay through stimulated emission, creating another photon. (4) The photons bounce back 
and forth through the laser medium between the mirrors, building up more and more photons. (5) A 
small percentage of the photons pass through the partially-silvered mirror to become the laser beam we 
see. 



309 

The basis of the laser is the laser material which consists of the atoms that are used to create the 
laser beam. Many different materials can be used as laser material, and their energy levels determine the 
characteristics of the laser. Some examples of different lasers are shown in Table 16.4. The laser material is 
contained in the optical cavity. 

Before the laser is turned on, all the atoms in the laser material are in their ground state. The first step 
in creating a laser beam is to add energy to the laser material to raise most of the electrons into an excited 
metastable state. This is called pumping the laser. 

The creation of the laser beam starts through the process oi spontaneous emission, shown in Figure 16.14. 
An electron drops down to the ground state and emits a photon with energy equal to the energy difference 
of the two energy levels. This laser photon is the beginning of the laser beam. 

At some time a laser photon will run into another excited electron. Then stimulated emission occurs 
and the electron drops down to the ground state and emits an additional identical photon as shown in 
Figure 16.15. Since the laser material typically has a large number of atoms, one laser photon passing 
through this material will rapidly cause a large number of photons just like it to be emitted. 

The optical cavity keeps the laser photons inside the laser cavity so that they can build up the laser 
beam. At each end is a concave mirror; one is a full mirror and one is a partial mirror. The full mirror is 
totally reflective. The partial mirror transmits a small amount of the light that hits it (less than 1%). The 
mirrors are carefully aligned so that photons that reflect off one mirror become "trapped", and bounce back 
and forth between the mirrors many times causing more and more stimulated emission. The photons that 
eventually escape through the partially-silvered mirror become the laser beam that we see. 

As the photons bounce between mirrors, they continually pass through the laser material, stimulating 
those atoms to emit more photons. This creates an ever increasing beam of photons, all with the same 
characteristics, all traveling in the same direction. In this way, the optical cavity helps to amplify the 
original laser photons into a concentrated, intense beam of photons. 

The laser cavity also helps to narrow the frequency range of laser light emitted. The distance between 
the two mirrors defines the cavity mode which only allows light of a narrow range of frequencies to continue 
being reflected back and forth. Light of other frequencies damped out. (This is just like in the chapter 
on the physics of music where a pipe of a certain length corresponds to a particular wavelength of sound.) 
Therefore only a narrow frequency of light can be emitted. 

NOTE: In 1953, Charles H. Townes and graduate students James P. Gordon and Herbert J. 
Zeiger produced the first maser, a device operating on similar principles to the laser, but producing 
microwave rather than optical radiation. Townes's maser was incapable of making a continuous 
beam. Nikolay Basov and Aleksandr Prokhorov of the former Soviet Union worked independently 
and developed a method of making a continuous beam using more than two energy levels. Townes, 
Basov and Prokhorov shared the Nobel Prize in Physics in 1964. 

16.4.1.3 Laser applications and safety 

Although the first working laser was only produced in 1958, lasers are now found in many household items. 
For example, lasers are well-known through their use as cheap laser pointers. However, lasers can be very 
dangerous to the human eye since a large amount of energy is focused into a very narrow beam. NEVER 
POINT A LASER POINTER INTO SOMEBODY'S EYES - IT CAN BLIND THEM FOR- 
EVER. 

Other uses include: 

• Semiconductor lasers which are small, efficient and cheap to make are used in CD players. 

• He-Ne Lasers are used in most grocery shops to read in the price of items using their barcodes. This 
makes the cashiers' job much quicker and easier. 

• High energy lasers are used in medicine as a cutting and welding tool. Eye surgery in particular make 
use of the precision of lasers to reattach the retinas of patients' eyes. The heat from cutting lasers also 
helps to stop the bleeding of a wound by burning the edges (called cauterising). 



310 CHAPTER 16. OPTICAL PHENOMENA AND PROPERTIES OF MATTER 



Image notjinished 



Figure 16.19 



run demo^ 



16.4.1.3.1 Case Study : Uses of lasers 



Do research in a library or on the Internet on one application of laser technology Explain how the technology 
works by using a laser. 

You will need to present your findings to the class in the form of a poster. You can think of any useful 
application, but to give you some ideas of where to start, some applications are listed below: 



• laser printers 

• laser communication and fibre optics 

• optical storage 

• using lasers as precision measurement tools 



• your own ideas.. 



16.4.1.3.2 Lasers 

1. Explain what is meant by spontaneous emission of radiation. 

2. Explain what is meant by stimulated emission of radiation. 

3. List the similarities and differences between spontaneous emission of radiation and stimulated emission 
of radiation. 

4. How is the light emitted by a laser different from the light emitted by a light bulb? 

5. Describe using a simple diagram, how a laser works. Your description should include the following 
concepts: metastable state and population inversion. 

6. Give examples of some materials that have been used for lasers. What do all these materials have in 
common? 

7. Describe how the laser cavity affects: 

• increasing amplification 

• concentrating beam intensity 

• narrowing the frequency of the beam 

8. List some applications of lasers. 

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<http://static.slidesharecdn.com/swf/ssplayer2.swf?doc=opticalphenomina-100513094656- 

phpapp01&stripped_title=optical-phenomina&userName=kwarne> 



Figure 16.20 



''http://phet. Colorado. edu/sims/lasers/lasersen.jnlp 



311 

16.4.2 Summary 

1. Light of the correct frequency can eject electrons from a metal. This is called the photoelectric effect. 

2. A metal has a work function which is the minimum energy needed to emit an electron from the metal. 

3. Emission spectra are formed by glowing gases. The pattern of the spectra is characteristic of the 
specific gas. 

4. Absorption spectra are formed when certain frequencies of light are absorbed by a material. 

5. Lasers are devices that produce a special type of light that has many uses. 

6. Lasers have many uses, for example, in CD and DVD players, to cut material, in surgery, in printing, 
in telecommunications and as laser pointers. 



16.4.3 End of chapter exercise 

1. What is the photoelectric effect? 

2. Calculate the energy of a photon of red light with a wavelength of 400 nm. 

3. Will ultraviolet light with a wavelenth of 990 nm be able to emit electrons from a sheet of calcium with 
a work function of 2,9 eV? 

4. What does the acronym LASER stand for? 

5. Name three types of lasers and their uses. 

6. Write a short essay on the benefits lasers have had on modern society. 



312 CHAPTER 16. OPTICAL PHENOMENA AND PROPERTIES OF MATTER 

Solutions to Exercises in Chapter 16 

Solution to Exercise 16.1 (p. 296) 

Step 1. We need to determine the maximum kinetic energy of an electron ejected from a silver foil by ultraviolet 
radiation. 



The photoelectric effect tells us that: 



^k — ^photon 

We also have: 

Work function of silver: ^ = 6, 9 x 10^^^ J 
UV radiation wavelength = 250 nm = 250 x 10~^ m 
Planck's constant: /i = 6, 63 x lO""^"* ni^kgs^^ 
speed of light: c = 3 x 10^ ins~^ 
Step 2. 

Ek = ^-4> 



(16.7) 



6, 63 X W~^^ X 250^10-8 ] - 6, 9 X 10^^^ (16.J 



1,06 X 10"^^ J 



The maximum kinetic energy of the emitted electron will be 1, 06 x 10 ^^ J. 

Solution to Exercise 16.2 (p. 297) 

For the electrons to be emitted from the surface, the energy of each photon needs to be greater than the 
work function of the material. 



Step 1. 



-^photon '^J 

= 6, 63 X lO"^'^ X 1, 2 X lO^'"^ (16.9) 

7,96x10"^^ J 
Therefore each photon of ultraviolet light has an energy of 7, 96 x 10~^^ J. 



Step 2. 
Step 3. 



(l)goid = 8,2x10-19 J (16.10) 

(16.11) 



7,96x10-19 J < 8,2x10-19 J 

-^photons ^ 'Pgold 

The energy of each photon is less than the work function of gold, therefore, the photons do not have 
enough energy to knock electrons out of the gold. No electrons would be emitted from the gold foil. 

Solution to Exercise 16.3 (p. 300) 

Step 1. We have an absorption line at 642 nm. This means that the substance in the glass container absorbed 
photons with a wavelength of 642 nm. We need to calculate which 2 energy levels of hydrogen this 
transition would correspond to. Therefore we need to know what energy the absorbed photons had. 

Step 2. 



E = f 



(6,63xl0--''*)x(3xl0*') 
642x10-3 

3,1 X 10-19 J 



(16.12) 



313 

The absorbed photons had energy of 3, 1 x 10^^^ 
Step 3. Figure 16.8 shows various energy level transitions. The transitions related to visible wavelengths are 
marked as the transitions beginning or ending on Energy Level 2. Let's find the energy of those 
transitions and compare with the energy of the absorbed photons we've just calculated. 
Energy of transition (absorption) from Energy Level 2 to Energy Level 3: 

£^2,3 = E2 — E3 

= 16,3x10^1^ J-19,4xl0~i^ J (16.13) 

-3,1 X 10"^^ J 

Therefore the energy of the photon that an electron must absorb to jump from Energy Level 2 to 
Energy Level 3 is 3, 1 x 10^^^ J. (NOTE: The minus sign means that absorption is occurring.) 
This is the same energy as the photons which were absorbed by the gas in the container! Therefore, 
since the transitions of all elements are unique, we can say that the gas in the container is hydrogen. 
The transition is absorption of a photon between Energy Level 2 and Energy Level 3. 

Solution to Exercise 16.4 (p. 301) 

Step 1. We are given A = 500 x 10^^ m and we need to find the photon's energy, frequency and colour. 
Step 2. 



E = !f 



(6,63xl0"-''*)x(3xl0*') 









500x10-" 




= 


3, 


98 X IQ-^^ J 




The energy of the photon is 3,98 x 10"^^ J. 






Step 3. 










E = 


= 


hf 




f = 


= 


E 
h 




_ 


_ 


3.98x10-'" 



(16.14) 



(16.15) 

~ 6,63xl0-"3* 

= 6 X lO^"* Hz 

The frequency of the photon is 6 x 10^'' Hz. 
Step 4. The wavelength given in the question is 500 nm. We can see in the table that green light has wavelengths 
between 492 - 577 nm. Therefore 500 nm is in this range so the colour of the light is green. 

Solution to Exercise 16.5 (p. 301) 

Step 1. Four wavelengths of light are given and we need to find their colours. 

We also need to find which colour photon has the highest energy and which one has the lowest energy. 
Step 2. a. 400 nm falls into the range for violet light (390 - 455 nm). 

b. 580 nm falls into the range for yellow light (577 - 597 nm). 

c. 650 nm falls into the range for red light (622 - 780 nm). 

d. 300 nm is not shown in the table. However, this wavelength is just a little shorter than the shortest 
wavelength in the violet range. Therefore 300 nm is ultraviolet. 

Step 3. We know E = !f 
For 400 nm: 

E = !f 

_ (6,63xl0-^^)x(3xl0'') ^g ^g^ 

~ 400x10-3 ^ ■ ' 

4,97 X 10"^^ J 



314 CHAPTER 16. OPTICAL PHENOMENA AND PROPERTIES OF MATTER 

For 580 nm: 



For 650 nm: 



For 300 nm: 



E 



E 



E 



he 
A 

(6,63xl0"^'')x(3xl0*') 
580x10-3 

3,43 X 10"^^ J 



he 
A 

(6,63xl0-^'')x(3xl0*') 
650x10-3 

3,06 X 10"^^ J 



he 
A 

(6,63xl0-^'')x(3xl0*') 
300x10-3 

6,63 X 10^1^ J 



Therefore, the photons with the highest energy are the ultraviolet photons. 
The photons with the lowest energy are from light which is red. 



(16.17) 



(16.18) 



(16.19) 



GLOSSARY 315 



Glossary 



A A reversible reaction 

A reversible reaction is a chemical reaction that can proceed in both the forward and reverse 
directions. In other words, the reactant and product of one reaction may reverse roles. 

Activation energy 

The energy that is needed to break the bonds in reactant molecules so that a chemical reaction 
can proceed. 

B Binary Number System 

A way of writing any number using only the digits and 1. 

Biological macromolecule 

A biological macromolecule is a polymer that occurs naturally in living organisms. These 
molecules are essential to the survival of life. 

Bit 

One bit is a short way of saying one 'binary digit'. It is a single or 1. 

C Catalyst 

A catalyst speeds up a chemical reaction, without being altered in any way. It increases the 
reaction rate by lowering the activation energy for a reaction. 

Chemical equilibrium 

Chemical equilibrium is the state of a chemical reaction, where the concentrations of the 
reactants and products have no net change over time. Usually, this state results when the 
forward chemical reactions proceed at the same rate as their reverse reactions. 

Collision theory 

Collision theory is a theory that explains how chemical reactions occur and why reaction rates 
differ for different reactions. The theory states that for a reaction to occur the reactant particles 
must collide, but that only a certain fraction of the total collisions, the effective collisions, 
actually cause the reactant molecules to change into products. This is because only a small 
number of the molecules have enough energy and the right orientation at the moment of impact 
to break the existing bonds and form new bonds. 

D De Broglie Hypothesis 

A particle of mass m moving with velocity v has a wavelength A related to is momentum p = mv 
by 

h h , ^ 

A=- = 12.5 

p mv 

This wavelength. A, is known as the de Broglie wavelength of the particle (where h is Planck's 
constant). 

Detergent 



316 GLOSSARY 

Detergents are compounds or mixtures of compounds that are used to assist cleaning. The term 
is often used to distinguish between soap and other chemical surfactants for cleaning. 

Diffraction 

Diffraction is the ability of a wave to spread out in wavefronts as the wave passes through a 
small aperture or around a sharp edge. 

Doppler Effect 

The Doppler effect is the apparent change in frequency and wavelength of a wave when the 
observer and the source of the wave move relative to each other. 

Doppler Effect 

when the wavelength and frequency measured by an observer are different to those emitted by 
the source due to movement of the source or observer. 

E Elastic Collisions 

An elastic collision is a collision where total momentum and total kinetic energy are both 
conserved. 

Elastic limit 

The elastic limit is the point beyond which permanent deformation takes place. 

Electrochemical reaction 

If a chemical reaction is caused by an external voltage, or if a voltage is caused by a chemical 
reaction, it is an electrochemical reaction. 

Electrode 

An electrode is an electrical conductor that is used to make contact with a metallic part of a 
circuit. The anode is the electrode where oxidation takes place. The cathode is the electrode 
where reduction takes place. 

Electrolysis 

In chemistry and manufacturing, electrolysis is a method of separating bonded elements and 
compounds by passing an electric current through them. 

Electrolyte 

An electrolyte is a substance that contains free ions and which therefore behaves as an electrical 
conductor. 

Electrolytic cell 

An electrolytic cell is a type of cell that uses electricity to drive a non-spontaneous reaction. 

Electromotive Force (emf) 

The emf of a cell is defined as the maximum potential difference between two electrodes or half 
cells in a voltaic cell, emf is the electrical driving force of the cell reaction. In other words, the 
higher the emf, the stronger the reaction. 

Equilibrium constant 

The equilibrium constant (Kc), relates to a chemical reaction at equilibrium. It can be calculated 
if the equilibrium concentration of each reactant and product in a reaction at equilibrium is 
known. 

Eutrophication 

Eutrophication refers to an increase in chemical nutrients in an ecosystem. These chemical 
nutrients usually contain nitrogen or phosphorus. 



GLOSSARY 317 

F Faraday's Law 

The emf (electromotive force), e, produced around a loop of conductor is proportional to the rate 
of change of the magnetic flux, (p, through the area, A, of the loop. This can be stated 
mathematically as: 

e = -N^ (13.1) 

At ^ ' 

where (j) = B ■ A and B is the strength of the magnetic field. 

Fertiliser 

A fertiliser is a compound that is given to a plant to promote growth. Fertilisers usually provide 
the three major plant nutrients and most are applied via the soil so that the nutrients are 
absorbed by plants through their roots. 

Fractional distillation 

Fractional distillation is the separation of a mixture into its component parts, or fractions. Since 
air is made up of a number of gases (with the major component being nitrogen), fractional 
distillation can be used to separate it into these different parts. 

Frame of Reference 

A frame of reference is the point of view from which a system is observed. 

Functional group 

In organic chemistry, a functional group is a specific group of atoms within molecules, that are 
responsible for the characteristic chemical reactions of those molecules. The same functional 
group will undergo the same or similar chemical reaction (s) regardless of the size of the molecule 
it is a part of. 

G Galvanic cell 

A galvanic (voltaic) cell is an electrochemical cell that uses a chemical reaction between two 
dissimilar electrodes dipped in an electrolyte, to generate an electric current. 

Generator 

A generator converts mechanical energy into electrical energy. 

H Half cell 

A half cell is a structure that consists of a conductive electrode surrounded by a conductive 
electrolyte. For example, a zinc half cell could consist of a zinc metal plate (the electrode) in a 
zinc sulphate solution (the electrolyte). 

Hooke's Law 

In an elastic spring, the extension varies linearly with the force applied. 

F = —kx 

where F is the restoring force in newtons (N), k is the spring constant in N ■ m~^ and x is the 

displacement of the spring from its equilibrium length in metres (m). 

Hydrocracking 

Hydrocracking is a cracking process that is assisted by the presence of an elevated partial 
pressure of hydrogen gas. It produces chemical products such as ethane, LPG, isoparafRns, jet 
fuel and diesel. 

I Impedance 



318 GLOSSARY 

The maximum voltage divided by the maximum current for any circuit. The unit of impedance 
is the ohm. 

Inelastic Collisions 

An inelastic collision is a collision in which total momentum is conserved but total kinetic 

energy is not conserved. The kinetic energy is transformed into other kinds of energy. 

Interference Minima 

The angle at which the minima in the interference spectrum occur is: 

sind= (11.1) 

a 

where 

is the angle to the minimum 
a is the width of the slit 

A is the wavelength of the impinging wavefronts 
m is the order of the mimimum, m = ±1, ±2, ±3, ... 
Isomer 

In chemistry, isomers are molecules with the same molecular formula and often with the same 
kinds of chemical bonds between atoms, but in which the atoms are arranged differently. 

J Joule 

1 joule is the work done when an object is moved 1 m under the application of a force of 1 N in 
the direction of motion. 

L Le Chatelier's Principle 

If a chemical system at equilibrium experiences a change in concentration, temperature or total 
pressure the equilibrium will shift in order to minimise that change and a new equilibrium is 
established. 

Limit of proportionality 

The limit of proportionality is the point beyond which Hooke's Law is no longer obeyed. 

M Mach Number 

The Mach Number is the ratio of the speed of an object to the speed of sound in the surrounding 
medium. 

Meta-stable state 

A meta-stable state is an excited atomic state that has an unusually long lifetime, compared to 
the lifetimes of other excited states of that atom. While most excited states have lifetimes 
measured in microseconds and nanoseconds (10^^ s and 10^^ s), meta-stable states can have 
lifetimes of milliseconds (10"'^ s) or even seconds. 

Modulo 

The modulo of a counter tells you how many stages (or pulses) it receives before going back to 
as its output. Thus a modulo 8 counter counts in eight stages 000, 001, 010, Oil, 100, 101, 110, 
111, then returns to 000 again. 

Motor 

An electric motor converts electrical energy into mechanical energy. 



GLOSSARY 319 

N Nutrient 

A nutrient is a substance that is used in an organism's metabolism or physiology and which must 
be taken in from the environment. 

O Open and closed systems 

An open system is one whose borders allow the movement of energy and matter into and out of 
the system. A closed system is one in which only energy can be exchanged, but not matter. 

P Permeability 

Permeability is the property of a material which describes the magnetisation developed in that 
material when excited by a source. 

Photon 

A photon is a quantum (energy packet) of light. 
Planck's constant 

Planck's constant is a physical constant named after Max Planck. 

/i = 6,626 X 10^3'' J • s 

Plastic 

Plastic covers a range of synthetic or semisynthetic organic polymers. Plastics may contain other 
substances to improve their performance. Their name comes from the fact that many of them 
are malleable, in other words they have the property of plasticity. 

Polymer 

Polymer is a term used to describe large molecules consisting of repeating structural units, or 
monomers, connected by covalent chemical bonds. 

Polymerisation 

In chemistry, polymerisation is a process of bonding monomers, or single units together through 
a variety of reaction mechanisms to form longer chains called polymers. 

Population inversion 

Population inversion is when more atoms are in an excited state than in their ground state. It is 
a necessary condition to sustain a laser beam, so that there are enough excited atoms that can 
be stimulated to emit more photons. 

Power 

Power is defined as the rate at which work is done or the rate at which energy is expended. The 
mathematical definition for power is: 

P = F-v (8.13) 

R Reactance 

The ratio of the maximum voltage to the maximum current when a capacitor or inductor is 
connected to an alternating voltage. The unit of reactance is the ohm. 

Reaction rate 

The rate of a reaction describes how quickly reactants are used up or how quickly products are 
formed during a chemical reaction. The units used are: moles per second (mols/second or 
mol.s"^). 

Resonance 



320 GLOSSARY 

Resonance occurs when a circuit is connected to an alternating voltage at its natural frequency. 
A very large current can build up in the circuit, even with minimal power input. 

S Salt bridge 

A salt bridge, in electrochemistry, is a laboratory device that is used to connect the oxidation 
and reduction half-cells of a galvanic cell. 

Soap 

Soap is a surfactant that is used with water for washing and cleaning. Soap is made by reacting 
a fat with either sodium hydroxide (NaOH) or potassium hydroxide (KOH). 

Sonic Boom 

A sonic boom is the sound heard by an observer as a Shockwave passes. 

Spontaneous Emission 

Spontaneous emission occurs when an atom is in an unstable excited state and randomly decays 
to a less energetic state, emitting a photon to carry off the excess energy. The unstable state 
decays in a characteristic time, called the lifetime. 

Standard emf (EcellO) 

Standard emf is the emf of a voltaic cell operating under standard conditions (i.e. 100 kPa, 
concentration = 1 mol.dm"^ and temperature = 298 K). The symbol " denotes standard 
conditions. 

Steam cracking 

Steam cracking occurs under very high temperatures. During the process, a liquid or gaseous 
hydrocarbon is diluted with steam and then briefly heated in a furnace at a temperature of 
about 850"^''C. Steam cracking is used to convert ethane to ethylene. Ethylene is a chemical 
that is needed to make plastics. Steam cracking is also used to make propylene, which is an 
important fuel gas. 

Stimulated emission 

Stimulated emission occurs when a photon interacts with an excited atom, causing the atom to 
decay and emit another identical photon. 

Subsonic 

Subsonic refers to speeds slower than the speed of sound. 
Supersonic 

Supersonic refers to speeds faster than the speed of sound. 

Surfactant 

A surfactant is a wetting agent that lowers the surface tension of a liquid, allowing it to spread 
more easily. 

T The Huygens Principle 

Each point on a wavefront acts like a point source of circular waves. The waves emitted from 
these point sources interfere to form another wavefront. 

The Lorentz Force 

The Lorentz force is the force experienced by a moving charged particle in a magnetic field and 
can be described by: 

F = qxvxB (13.2) 



GLOSSARY 321 

where 

F is the force (in newtons, N) 
q is the electric charge (in coulombs, C) 
V is the velocity of the charged particle (in m.s~^) and 
B is the magnetic field strength (in teslas, T). 
The photoelectric effect 

The photoelectric effect is the process whereby an electron is emitted by a metal when light 
shines on it. 

W Work 

When a force exerted on an object causes it to move, work is done on the object (except if the 
force and displacement are at right angles to each other). 

Work-Energy Theorem 

The work-energy theorem states that the work done on an object is equal to the change in its 
kinetic energy: 

W = AKE = KEf - KE, (8.8) 



322 



INDEX 



Index of Keywords and Terms 

Keywords are listed by the section with that keyword (page numbers are in parentheses). Keywords 
do not necessarily appear in the text of the page. They are merely associated with that section. Ex. 
apples, § 1.1 (1) Terms are referenced by the page they appear on. Ex. apples, 1 



2D wavefronts, 
§ 11.3(225) 

3D wavefronts, 
§ 11.3(225) 



11.1(219), § 11.2(222), 



11.1(219), § 11.2(222), 



A A reversible reaction, 66 

absorption spectra, § 16.3(298) 
Activation energy, 62 
alcohols, § 1.3(19) 
alternating current, § 13.2(248) 
amine, § 1.4(21) 
applications, § 4.5(97) 

B balancing redox, § 4.4(96) 
batteries, § 5.4(123) 
Binary Number System, 266 
bioligical macromolecules, § 2.2(43) 
Biological macromolecule, 43 
Bit, 269 

C capacitance, § 13.2(248) 

capacitive circuits, § 14.1(255) 

carboxylic acid, § 1.4(21) 

Catalyst, 63 

chemical equilibrium, § 3.3(64), 67, § 3.4(67), 

§ 3.5(70) 

chemical industry, § 5.1(107), § 5.2(113), 

§ 5.3(118), § 5.4(123) 

chemistry § 1.1(1), § 1.2(7), § 1.3(19), 

§ 1.4(21), § 2.1(33), § 2.2(43), § 3.1(53), 

§ 3.2(58), § 3.3(64), § 3.4(67), § 3.5(70), 

§ 4.1(81), § 4.2(87), § 4.3(89), § 4.4(96), 

§ 4.5(97), § 5.1(107), § 5.2(113), § 5.3(118), 

§ 5.4(123) 

chloralkali industry, § 5.2(113) 

circuit elements, § 14.3(271) 

collision theory, § 3.2(58), 58 

colour, § 10.1(209), § 10.2(214) 

conservation of momentum, § 6.2(137) 

D De Broglie Hypothesis, 236 

de Broglie wavelength, § 12.1(235) 



deformation, § 7.1(171) 
Detergent, 117 
diffraction, § 11.2(222), 222 
digital electronics, § 14.2(260) 
Doppler Effect, § 9(201), 201, 226 

E Elastic Collisions, 139 
Elastic Hmit, 174 

electrochemical reaction, § 4.1(81), 82 
electrochemical reactions, § 4.5(97) 
electrochemistry § 4.1(81), § 4.2(87), 
§ 4.3(89), § 4.4(96), § 4.5(97), § 5.4(123) 
Electrode, 84 

electrodynamics, § 13.1(243), § 13.2(248) 
Electrolysis, 87 
Electrolyte, 85 
electrolytic cell, § 4.2(87), 87 
Electromotive Force (emf), 93 
electron microscope, § 12.2(237) 
electronics, § 14.1(255), § 14.2(260), 
§ 14.3(271) 

EM radiation, § 15.1(283), § 15.2(287) 
emission spectra, § 16.3(298) 
energy § 8.2(185) 
equilibrium constant, § 3.4(67), 67 
ester, § 1.4(21) 
Eutrophication, 122 

F failure, § 7.2(175) 
Faraday's Law, 244 
Fertiliser, 120 

fertilizer industry, § 5.3(118) 
fertilizers, § 5.3(118) 
Fractional distillation, 110 
Frame of Reference, 144 
frames of reference, § 6.4(144) 
Functional group, 5 

G galvanic cell, § 4.1(81), 85 
Generator, 244 
generators, § 13.1(243) 
grade 12, § 1.1(1), § 1.2(7), § 1.3(19), 



INDEX 



323 



1-4(21), § 


2.1(33), § 2.2(43), § 3.1(53), 


3.2(58), § 3.3(64), § 3.4(67), § 3.5(70), 


4-1(81), § 


4.2(87), § 4.3(89), § 4.4(96), 


4-5(97), § 


5.1(107), § 5.2(113), § 5.3(118), 


5.4(123), ! 


5 6.1(133), § 6.2(137), § 6.3(139), 


6.4(144), 1 


5 7.1(171), § 7.2(175), § 8.1(181), 


8.2(185), ! 


5 8.3(190), § 9(201), § 10.1(209), 


10.2(214), 


§ 11.1(219), § 11.2(222), 


11.3(225), 


§ 12.1(235), § 12.2(237), 


13.1(243), 


§ 13.2(248), § 14.1(255), 


14.2(260), 


§ 14.3(271), § 15.1(283), 


15.2(287), 


§ 16.1(291), § 16.2(294), 


16.3(298), 


§ 16.4(303) 



o 



H Half cell, 84 

Hooke's law, § 7.1(171), 171 
Huygen's principle, § 11.1(219) 
hydrocarbons, § 1.2(7) 
Hydrocracking, 109 

I Impedance, 257 

inductance, § 13.2(248) 
inductive circuits, § 14.1(255) 
Inelastic Collisions, 141 
interference, § 11.1(219) 
Interference Minima, 225 
introduction, § 1.1(1), § 3.1(53) 
Isomer, 4 

J Joule, 182 

L lasers, § 16.4(303) 

Le Chatelier's principle, § 3.5(70), 70 
light, § 10.1(209), § 16.1(291) 
Limit of proportionality, 174 
logic gates, § 14.2(260) 

M Mach Number, 227 

materials, § 7.1(171), § 7.2(175) 

matter, § 7.1(171), § 7.2(175), § 12.1(235), 

§ 12.2(237) 

measurement, § 3.2(58) 

mechanical properties, § 7.1(171), § 7.2(175) 

mechanism, § 3.2(58) 

Meta-stable state, 304 

Modulo, 268 

motion, § 6.1(133), § 6.2(137), § 6.3(139), 

§ 6.4(144) 

Motor, 246 

motors, § 13.1(243) 

N Nutrient, 118 



Open and closed systems, 65 

optical phenomena, § 16.1(291), § 16.2(294), 

§ 16.3(298), § 16.4(303) 

organic macromolecules, § 2.1(33), § 2.2(43) 

organic molecules, § 1.1(1), § 1.2(7), § 1.3(19), 

§ 1-4(21) 

paints, § 10.2(214) 

particle nature, § 15.1(283) 

penetrating ability, § 15.2(287) 

Permeability, 250 

photoelectric effect. 

Photon, 283 

physics, § 6.1(133), 

§ 6.4(144), § 7.1(171), § 

§ 8.2(185), § 8.3(190), § 

§ 10.2(214), § 11.1(219) 

§ 11.3(225), § 12.1(235) 

§ 13.1(243), § 13.2(248) 

§ 14.2(260), § 14.3(271) 

§ 15.2(287), § 16.1(291) 

§ 16.3(298), § 16.4(303) 

pigments, § 10.2(214) 

Planck's constant, 286 

Plastic, 38 

Polymer, 33 

Polymerisation, 34 

polymers, § 2.1(33) 

Population inversion, 306 

power, § 8.3(190), 190 

properties of matter, § 16.1(291), § 16.2(294), 

§ 16.3(298), § 16.4(303) 



R 



16.2(294) 

6.2(137), § 6.3(139), 
, § 7.2(175), § 8.1(181), 
, § 9(201), § 10.1(209), 

11.2(222), 

12.2(237), 

14.1(255), 

15.1(283), 

16.2(294), 



3.2(58), § 3.3(64), 



Reactance, 256 
Reaction rate, 54 
reaction rates, § 3.1(53), 
§ 3.4(67), § 3.5(70) 
redox reaction, § 4.4(96) 
Resonance, 259 



Salt bridge, 85 

sasol, § 5.1(107) 

scattering, § 16.1(291) 

science, § 1.1(1), § 1.2(7), § 1.3(19), § 1.4(21), 

§ 2.1(33), § 2.2(43), § 3.1(53), § 3.2(58), 
3.5(70), § 4.1(81), 
4.4(96), § 4.5(97), 
, § 5.3(118), § 5.4(123), 



3.3(64), § 
4-2(87), § 
5.1(107), 
6.1(133), 
7.1(171), 
8.3(190), 
11.1(219), 



3.4(67), 
4.3(89), 
5 5.2(113), § 
5 6.2(137), § 
5 7.2(175), § 
5 10.1(209), 
§ 11.2(222) 



6.3(139), § 6.4(144), 
8.1(181), § 8.2(185), 
5 10.2(214), 
§ 11-3(225), 



324 



INDEX 



§ 12.1(235), § 12.2(237), § 13.1(243), 

§ 13.2(248), § 14.1(255), § 14.2(260), 

§ 14.3(271), § 15.1(283), § 15.2(287), 

§ 16.1(291), § 16.2(294), § 16.3(298), 

§ 16.4(303) 

shock waves, § 11.3(225) 

Soap, 117 

Sonic Boom, 227 

sonic booms, § 11.3(225) 

South Africa, § 1.1(1), § 1.2(7), § 1.3(19), 

§ 1.4(21), § 2.1(33), § 2.2(43), § 3.1(53), 

§ 3.2(58), § 3.3(64), § 3.4(67), § 3.5(70), 

§ 4.1(81), § 4.2(87), § 4.3(89), § 4.4(96), 

§ 4.5(97), § 5.1(107), § 5.2(113), § 5.3(118), 

§ 5.4(123), § 6.1(133), § 6.2(137), § 6.3(139), 

§ 6.4(144), § 7.1(171), § 7.2(175), § 8.1(181), 

§ 8.2(185), § 8.3(190), § 9(201), § 10.1(209), 

§ 10.2(214), § 11.1(219), § 11.2(222), 

§ 11.3(225), § 12.1(235), § 12.2(237), 

§ 13.1(243), § 13.2(248), § 14.1(255), 

§ 14.2(260), § 14.3(271), § 15.1(283), 

§ 15.2(287), § 16.1(291), § 16.2(294), 

§ 16.3(298), § 16.4(303) 

Spontaneous Emission, 304 



Standard emf (EcellO), 93 
standard potential, § 4.3(89) 
Steam cracking, 109 
Stimulated emission, 304 
strength, § 7.2(175) 
Subsonic, 226 
Supersonic, 226 
Surfactant, 116 

T The Huygens Principle, 220 
The Lorentz Force, 246 
The photoelectric effect, 294 
transmission, § 16.1(291) 
two-dimensions, § 6.1(133), § 6.2(137), 
§ 6.4(144) 

two-dimenstions, § 6.3(139) 
types of collisions, § 6.3(139) 

V verticle projectiles, § 6.1(133) 

W wave nature, § 12.1(235), § 12.2(237), 
§ 15.1(283) 
work, § 8.1(181), 181 
Work-Energy Theorem, 187 



ATTRIBUTIONS 325 

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Used here as: "Chemical equilibrium" 

By: Free High School Science Texts Project 

URL: http://siyavula.cnx.Org/content/m39469/l.l/ 

Pages: 64-67 

Copyright: Free High School Science Texts Project 

License: http://creativecommons.Org/licenses/by/3.0/ 

Module: "Reaction Rates: the equilibrium constant" 

Used here as: "The equilibrium constant" 

By: Free High School Science Texts Project 

URL: http://siyavula.cnx.Org/content/m39473/l.l/ 

Pages: 67-70 

Copyright: Free High School Science Texts Project 

License: http://creativecommons.Org/licenses/by/3.0/ 

Module: "Reaction Rates: Le Chateliers principle" 

Used here as: "Le Chateliers principle" 

By: Free High School Science Texts Project 

URL: http://siyavula.cnx.Org/content/m39475/l.l/ 

Pages: 70-77 

Copyright: Free High School Science Texts Project 

License: http://creativecommons.Org/licenses/by/3.0/ 

Module: "Electrochemical Reactions: The galvanic cell" 

Used here as: "The galvanic cell" 

By: Free High School Science Texts Project 

URL: http://siyavula.cnx.Org/content/m39417/l.l/ 

Pages: 81-86 

Copyright: Free High School Science Texts Project 

License: http://creativecommons.Org/licenses/by/3.0/ 



ATTRIBUTIONS 327 

Module: "Electrochemical Reactions: The electrolytic cell" 

Used here as: "The electrolytic cell" 

By: Free High School Science Texts Project 

URL: http://siyavula.cnx.Org/content/m39419/l.l/ 

Pages: 87-89 

Copyright: Free High School Science Texts Project 

License: http://creativecommons.Org/licenses/by/3.0/ 

Module: "Electrochemical Reactions: Standard potentials" 

Used here as: "Standard potentials" 

By: Free High School Science Texts Project 

URL: http://siyavula.cnx.Org/content/m39420/l.l/ 

Pages: 89-96 

Copyright: Free High School Science Texts Project 

License: http://creativecommons.Org/licenses/by/3.0/ 

Module: "Electrochemical Reactions: Balancing redox reactions" 

Used here as: "Balancing redox reactions" 

By: Free High School Science Texts Project 

URL: http://siyavula.cnx.Org/content/m39422/l.l/ 

Pages: 96-97 

Copyright: Free High School Science Texts Project 

License: http://creativecommons.Org/licenses/by/3.0/ 

Module: "Electrochemical Reactions: Applications" 

Used here as: "Applications" 

By: Free High School Science Texts Project 

URL: http://siyavula.cnx.Org/content/m39424/l.l/ 

Pages: 97-102 

Copyright: Free High School Science Texts Project 

License: http://creativecommons.Org/licenses/by/3.0/ 

Module: "The chemical industry: Sasol" 

Used here as: "Sasol" 

By: Free High School Science Texts Project 

URL: http://siyavula.cnx.Org/content/m39484/l.l/ 

Pages: 107-113 

Copyright: Free High School Science Texts Project 

License: http://creativecommons.Org/licenses/by/3.0/ 

Module: "The chemical industry: the chloralkali industry" 

Used here as: "The chloralkali industry" 

By: Free High School Science Texts Project 

URL: http://siyavula.cnx.Org/content/m39479/l.l/ 

Pages: 113-118 

Copyright: Free High School Science Texts Project 

License: http://creativecommons.Org/licenses/by/3.0/ 



328 ATTRIBUTIONS 

Module: "The chemical industry: the fertilizer industry" 

Used here as: "The fertilizer industry" 

By: Free High School Science Texts Project 

URL: http://siyavula.cnx.Org/content/m39482/l.l/ 

Pages: 118-123 

Copyright: Free High School Science Texts Project 

License: http://creativecommons.Org/licenses/by/3.0/ 

Module: "The chemical industry: electrochemistry and batteries" 

Used here as: "Electrochemistry and batteries" 

By: Free High School Science Texts Project 

URL: http://siyavula.cnx.Org/content/m39485/l.l/ 

Pages: 123-131 

Copyright: Free High School Science Texts Project 

License: http://creativecommons.Org/licenses/by/3.0/ 

Module: "Motion in two dimensions: vertical projectile motion" 

Used here as: "Vertical projectile motion" 

By: Free High School Science Texts Project 

URL: http://siyavula.cnx.Org/content/m39546/l.l/ 

Pages: 133-137 

Copyright: Free High School Science Texts Project 

License: http://creativecommons.Org/licenses/by/3.0/ 

Module: "Motion in two dimensions: Conservation of momentum" 

Used here as: "Conservation of momentum" 

By: Free High School Science Texts Project 

URL: http://siyavula.cnx.Org/content/m39550/l.l/ 

Pages: 137-139 

Copyright: Free High School Science Texts Project 

License: http://creativecommons.Org/licenses/by/3.0/ 

Module: "Motion in two dimensions: Types of collisions" 

Used here as: "Types of collisions" 

By: Free High School Science Texts Project 

URL: http://siyavula.cnx.Org/content/m39539/l.l/ 

Pages: 139-144 

Copyright: Free High School Science Texts Project 

License: http://creativecommons.Org/licenses/by/3.0/ 

Module: "Motion in two dimensions: Frames of reference" 

Used here as: "Frames of reference" 

By: Free High School Science Texts Project 

URL: http://siyavula.cnx.Org/content/m39542/l.l/ 

Pages: 144-154 

Copyright: Free High School Science Texts Project 

License: http://creativecommons.Org/licenses/by/3.0/ 



ATTRIBUTIONS 329 

Module: "Mechanical Properties of Matter: deformation of materials" 

Used here as: "Deformation of materials" 

By: Free High School Science Texts Project 

URL: http://siyavula.cnx.Org/content/m39533/l.l/ 

Pages: 171-174 

Copyright: Free High School Science Texts Project 

License: http://creativecommons.Org/licenses/by/3.0/ 

Module: "Mechanical Properties of Matter: failure and strength of materials" 

Used here as: "Failure and strength of materials" 

By: Free High School Science Texts Project 

URL: http://siyavula.cnx.Org/content/m39537/l.l/ 

Pages: 175-178 

Copyright: Free High School Science Texts Project 

License: http://creativecommons.Org/licenses/by/3.0/ 

Module: "Work (Grade 12)" 

Used here as: "Work" 

By: Free High School Science Texts Project 

URL: http://siyavula.cnx.Org/content/m39594/l.l/ 

Pages: 181-185 

Copyright: Free High School Science Texts Project 

License: http://creativecommons.Org/licenses/by/3.0/ 

Module: "Energy (Grade 12)" 

Used here as: "Energy" 

By: Free High School Science Texts Project 

URL: http://siyavula.cnx.Org/content/m39592/l.l/ 

Pages: 185-190 

Copyright: Free High School Science Texts Project 

License: http://creativecommons.Org/licenses/by/3.0/ 

Module: "Power (Grade 12)" 

Used here as: "Power" 

By: Free High School Science Texts Project 

URL: http://siyavula.cnx.Org/content/m39598/l.l/ 

Pages: 190-194 

Copyright: Free High School Science Texts Project 

License: http://creativecommons.Org/licenses/by/3.0/ 

Module: "Doppler Effect" 

By: Rory Adams, Free High School Science Texts Project, Kosma von Maltitz, Heather Williams 

URL: http://siyavula.cnx.Org/content/m30847/l.4/ 

Pages: 201-207 

Copyright: Rory Adams, Free High School Science Texts Project, Heather Williams 

License: http://creativecommons.Org/licenses/by/3.0/ 

Module: "Colour: colour and light" 

Used here as: "Colour and light" 

By: Free High School Science Texts Project 

URL: http://siyavula.cnx.Org/content/m39506/l.l/ 

Pages: 209-214 

Copyright: Free High School Science Texts Project 

License: http://creativecommons.Org/licenses/by/3.0/ 



330 ATTRIBUTIONS 

Module: "Colour: paints and pigments" 

Used here as: "Paints and pigments" 

By: Free High School Science Texts Project 

URL: http://siyavula.cnx.Org/content/m39505/l.l/ 

Pages: 214-216 

Copyright: Free High School Science Texts Project 

License: http://creativecommons.Org/licenses/by/3.0/ 

Module: "2D and 3D Wavefronts: wavefronts, Huygen's principle and interference" 

Used here as: "Wavefronts, Huygen's principle and interference" 

By: Free High School Science Texts Project 

URL: http://siyavula.cnx.Org/content/m39488/l.l/ 

Pages: 219-222 

Copyright: Free High School Science Texts Project 

License: http://creativecommons.Org/licenses/by/3.0/ 

Module: "2D and 3D Wavefronts: Diffraction" 

Used here as: "Diffraction" 

By: Free High School Science Texts Project 

URL: http://siyavula.cnx.Org/content/m39491/l.l/ 

Pages: 222-225 

Copyright: Free High School Science Texts Project 

License: http://creativecommons.Org/licenses/by/3.0/ 

Module: "2D and 3D Wavefronts: Shock waves and sonic booms" 

Used here as: "Shock waves and sonic booms" 

By: Free High School Science Texts Project 

URL: http://siyavula.cnx.Org/content/m39503/l.l/ 

Pages: 225-231 

Copyright: Free High School Science Texts Project 

License: http://creativecommons.Org/licenses/by/3.0/ 

Module: "Wave nature of matter: de Broglie wavelength" 

Used here as: "de Broglie wavelength" 

By: Free High School Science Texts Project 

URL: http://siyavula.cnx.Org/content/m39589/l.l/ 

Pages: 235-236 

Copyright: Free High School Science Texts Project 

License: http://creativecommons.Org/licenses/by/3.0/ 

Module: "Wave nature of matter: Electron microscopes" 

Used here as: "Electron microscopes" 

By: Free High School Science Texts Project 

URL: http://siyavula.cnx.Org/content/m39574/l.l/ 

Pages: 237-239 

Copyright: Free High School Science Texts Project 

License: http://creativecommons.Org/licenses/by/3.0/ 



ATTRIBUTIONS 331 

Module: "Electrodynamics: Generators and motors" 

Used here as: "Generators and motors" 

By: Free High School Science Texts Project 

URL: http://siyavula.cnx.Org/content/m39508/l.l/ 

Pages: 243-248 

Copyright: Free High School Science Texts Project 

License: http://creativecommons.Org/licenses/by/3.0/ 

Module: "Electrodynamics: Alternating current, inductance and capacitance" 

Used here as: "Alternating current, inductance and capacitance" 

By: Free High School Science Texts Project 

URL: http://siyavula.cnx.Org/content/m39510/l.l/ 

Pages: 248-252 

Copyright: Free High School Science Texts Project 

License: http://creativecommons.Org/licenses/by/3.0/ 

Module: "Electronics: capacitive and inductive circuits" 

Used here as: "Capacitive and inductive circuits" 

By: Free High School Science Texts Project 

URL: http://siyavula.cnx.Org/content/m39523/l.l/ 

Pages: 255-260 

Copyright: Free High School Science Texts Project 

License: http://creativecommons.Org/licenses/by/3.0/ 

Module: "Electronics: principles of digital electronics" 

Used here as: "Principles of digital electronics" 

By: Free High School Science Texts Project 

URL: http://siyavula.cnx.Org/content/m39530/l.l/ 

Pages: 260-271 

Copyright: Free High School Science Texts Project 

License: http://creativecommons.Org/licenses/by/3.0/ 

Module: "Electronics: active circuit elements" 

Used here as: "Active circuit elements" 

By: Free High School Science Texts Project 

URL: http://siyavula.cnx.Org/content/m39531/l.l/ 

Pages: 271-280 

Copyright: Free High School Science Texts Project 

License: http://creativecommons.Org/licenses/by/3.0/ 

Module: "EM radiation: wave nature and particle nature (Grade 12)" 

Used here as: "Wave nature and particle nature" 

By: Free High School Science Texts Project 

URL: http://siyavula.cnx.Org/content/m39511/l.l/ 

Pages: 283-286 

Copyright: Free High School Science Texts Project 

License: http://creativecommons.Org/licenses/by/3.0/ 



332 ATTRIBUTIONS 

Module: "EM radiation: penetrating ability (Grade 12)" 

Used here as: "Penetrating ability" 

By: Free High School Science Texts Project 

URL: http://siyavula.cnx.Org/content/m39513/l.l/ 

Pages: 287-289 

Copyright: Free High School Science Texts Project 

License: http://creativecommons.Org/licenses/by/3.0/ 

Module: "Optical Phenomena and Properties of Matter: transmission and scattering of light" 

Used here as: "Transmission and scattering of light" 

By: Free High School Science Texts Project 

URL: http://siyavula.cnx.Org/content/m39560/l.l/ 

Pages: 291-294 

Copyright: Free High School Science Texts Project 

License: http://creativecommons.Org/licenses/by/3.0/ 

Module: "Optical Phenomena and Properties of Matter: photoelectric effect" 

Used here as: "Photoelectric effect" 

By: Free High School Science Texts Project 

URL: http://siyavula.cnx.Org/content/m39551/l.l/ 

Pages: 294-297 

Copyright: Free High School Science Texts Project 

License: http://creativecommons.Org/licenses/by/3.0/ 

Module: "Optical Phenomena and Properties of Matter: emission and absorption spectra" 

Used here as: "Emission and absorption spectra" 

By: Free High School Science Texts Project 

URL: http://siyavula.cnx.Org/content/m39553/l.l/ 

Pages: 298-303 

Copyright: Free High School Science Texts Project 

License: http://creativecommons.Org/licenses/by/3.0/ 

Module: "Optical Phenomena and Properties of Matter: lasers" 

Used here as: "Lasers" 

By: Free High School Science Texts Project 

URL: http://siyavula.cnx.Org/content/m39557/l.l/ 

Pages: 303-311 

Copyright: Free High School Science Texts Project 

License: http://creativecommons.Org/licenses/by/3.0/ 



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